<Parag>Here you can find intermediate algebra lessons that will help you understand by explaining in detail and clarifying with some examples.</Parag>

Here you can find intermediate algebra lessons that will help you understand by explaining in detail and clarifying with some examples.

Intermediate Algebra

# Arithmetic Sequence (Progression): Definition and Examples

Look closely at this series: <Tex>3;\\:7;\\:11;\\:15;\\:19;...</Tex>

what do you notice?

It is very clear that there is some relationship between each number that follows it or the one before it. This relationship is that the difference between every two **consecutive** numbers is 4. That is, if we know the value of the term in the current position, we can know the value of the term in the next or previous position only by adding the number 4 to the value of the term in the current position.

<Image src="/imgs/lessons/intermalgebra/ar-seq.png" alt="arithmetic sequence formula" width={866} height={233} />

This fixed number is called **common difference** and we denote it by d.

The number d determines the next term by adding d to the current term, and so on. This is how arithmetic sequences are generated.

There are three different cases of the number d:

- If d > 0 then the sequence is increasing, that is, each term is greater than the term before it.
- If d < 0 then the sequence is decreasing, that is, each term is less than the term before it.
- If d = 0, then the sequence is constant.

<Note>

An arithmetic sequence (i.e arithmetic progression) is in which the difference between each term the preceding term is always constant.

</Note>

## Practical examples of arithmetic sequences

<Example type="example" num="1"/>

Find the missing number in the following arithmetic sequence:

<Tex block>1\\:;\\:6\\:;\\:11\\:;\\:16\\:;\\:?</Tex>

<Example type="solution"/>

First we must calculate d (common difference)

<Tex block>
\begin{align*}
6-1&=5\\\
11-6&=5\\\
16-11&=5
\end{align*}
</Tex>

In these variances we got a fixed number. So <Tex>d=5</Tex>, which means that the sequence is increasing.

Now, in order to find the missing number, it is enough to add 5 to the number before the number we are looking for which is 16.

<Tex block>16+5=21</Tex>

So the arithmetic sequence is:

<Tex block>1\\:;\\:6\\:;\\:11\\:;\\:16\\:;\\:21</Tex>

-------------

<Example type="example" num="2"/>

Find the missing number in the following arithmetic sequence:

<Tex block>
5\\:;\\:1\\:;\\:-3\\:;\\:-7\\:;\\:?
</Tex>

<Example type="solution"/>

First we must calculate d (common difference):

<Tex block>
\begin{align*}
1-5&=-4\\\
-3-1&=-4\\\
-7-(-3)&=-4
\end{align*}
</Tex>

In these variances we got a fixed number. So <Tex>d=-4</Tex>, which means that the sequence is decreasing.

Now, in order to find the missing number, it is enough to add -4 to the number before the number we are looking for which is -7.

<Tex block>-7+(-4)=-11</Tex>

So the arithmetic sequence is:

<Tex block>5\\:;\\:1\\:;\\:-3\\:;\\:-7\\:;\\:-11</Tex>

-------------

<Example type="example" num="3"/>

Find the missing number in the following arithmetic sequence:

<Tex block>?\\:;\\:8\\:;\\:14\\:;\\:20\\:;\\:26</Tex>

<Example type="solution"/>

First we must calculate d (common difference)

<Tex block>
\begin{align*}
14-8&=6\\\
20-14&=6\\\
26-20&=6
\end{align*}
</Tex>

In these variances we got a fixed number. So <Tex>d=6</Tex>, which means that the sequence is increasing.

Now, in order to find the missing number, it is enough to subtract 6 from the number after the number we are looking for which is 8.

<Tex block>8-6=2</Tex>

So the arithmetic sequence is:

<Tex block>2\\:;\\:8\\:;\\:14\\:;\\:20\\:;\\:26</Tex>

-------------

<Example type="example" num="4"/>

Find the missing numbers in the following series for which they are an arithmetic sequence:

<Tex block>6\\:;\\:?\\:;\\:?\\:;\\:?\\:;\\:34</Tex>

<Example type="solution"/>

In this case we should look for a number that will give us a steady pace to start from 6 and get to 34.

The method of calculating d in this case:

<Note>
<Tex block>d=\frac{a_m-a_n}{k+1}</Tex>
- In which <Tex>a_m</Tex> and <Tex>a_n</Tex> two different terms of an arithmetic sequence.
- The number <Tex>k</Tex> represents the number of missing numbers between <Tex>a_m</Tex> and <Tex>a_n</Tex>.
</Note>

We have in the example:

<Tex block>a_m=34\quad;\quad a_n=6\quad;\quad k=3</Tex>

Then,

<Tex block>
\begin{align*}
d&=\frac{34-6}{3+1}\\\
d&=\frac{28}{4}\\\
d&=7
\end{align*}
</Tex>

So the arithmetic sequence is:

<Tex block>6\\:;\\:13\\:;\\:20\\:;\\:27\\:;\\:34</Tex>

this lesson show you definition of arithmetic sequence (progression sequence) and the common difference with practical examples of increasing and decreasing sequences.

Arithmetic Sequence: Definition and Examples

# Arithmetic Sequence Formula: How to Find General Term

<Related>

<Link to="../intermediate-algebra/arithmetic-sequence-definition-and-examples" as="/lessons/intermediate-algebra/arithmetic-sequence-definition-and-examples">Arithmetic Sequence (Progression): Definition and Examples</Link>

</Related>


We saw in the definition of an arithmetic sequence that we can write the following term in terms of the current term according to the following formula:

<Tex block>a_{n+1}=a_n+d</Tex>

so that: <Tex>a_{n+1}</Tex> is the next term;

and <Tex>a_{n}</Tex> is the current term;

and <Tex>d</Tex> is the common difference.

So, as you notice through this formula that one of the two terms is written in terms of the other. But, can we write the term <Tex>a_{n}</Tex> in terms of <Tex>n</Tex> only?

This is what we will answer in this lesson.

Let's take the previous formula <Tex>a_{n+1}=a_n+d</Tex> and do some calculations:

<Tex block>
\begin{align*}
a_1&=a_1 \\\
a_2&=a_1+d \\\
a_3&=a_2+d=(a_1+d)+d=a_1+2d \\\
a_4&=a_3+d=(a_1+2d)+d=a_1+3d\\\
a_5&=a_4+d=(a_1+3d)+d=a_1+4d \\\
&\quad \vdots \quad \vdots \\\
&\quad \vdots \quad \vdots \\\
a_n&=a_1+(n-1)d
\end{align*}
</Tex>

This formula that we get is called the general term of an arithmetic sequence. Through it we can calculate the value of any term in position <Tex>n</Tex> directly.

<Note>

Let <Tex>(a_n)</Tex> an arithmetic sequence with common difference <Tex>d</Tex>. The term general of <Tex>(a_n)</Tex> is:
<Tex block>a_n=a_1+(n-1)d</Tex>


</Note>

Let's take some examples

------------------

<Example type="example" num="1"/>

We consider the arithmetic sequence: <Tex>-3;\\; 2; \\; 7; \\; 12; \\; \ldots</Tex>

Find 20th term

<Example type="solution"/>

First, we must find <Tex>d</Tex>

<Tex>d=2-(-3)=7-2=12-7=5</Tex>

Thus,

<Tex block>
\begin{align*}
a_n &=a_1+(n-1)d \\\
a_{20} &=a_1+(20-1)\times5 \\\
a_{20} &=-3+19\times5 \\\
a_{20} &=-3+95 \\\
a_{20} &=92
\end{align*}
</Tex>

------------------

<Example type="example" num="2"/>

Let <Tex>(a_n)</Tex> an arithmetic sequence with <Tex>a_1 =4</Tex> and <Tex>a_{11} = 34</Tex>. Find <Tex>a_{200}</Tex>

<Example type="solution"/>

First, we must find <Tex>d</Tex>

<Tex block>
\begin{align*}
a_n &=a_1+(n-1)d \\\
a_{11} &=a_1+(11-1)\times d \\\
34 &=4+10\times d \\\
34 &=4+10d \\\
10d &=34-4=30 \\\
d &= \frac{30}{10}=3
\end{align*}
</Tex>

Thus,

<Tex block>
\begin{align*}
a_n &=a_1+(n-1)d \\\
a_{200} &=a_1+(200-1)\times3 \\\
a_{200} &=4+199\times3 \\\
a_{200} &=4+597 \\\
a_{200} &=601
\end{align*}
</Tex>

------------------

<Example type="example" num="3"/>

Et <Tex>(a_n)</Tex> an arithmetic sequence with <Tex>a_1=3</Tex> and <Tex>d=4</Tex>.

Is 59 a term belongs to the arithmetic sequence <Tex>(a_n)</Tex>?

<Example type="solution"/>

In order 59 to be a term to belong to the arithmetic sequence <Tex>(a_n)</Tex>, it must satisfy the general term formula such that <Tex>n</Tex> is a natural integer.

We have:

<Tex block>
\begin{align*}
a_n &= a_1+(n-1)d \\\
59 &=3+(n-1)\times4 \\\
59 &=3+4n-4 \\\
59 &=4n-1 \\\
4n &=60 \\\
n &= \frac{60}{4} \\\
n &= 15
\end{align*}
</Tex>

Since 15 is a natural number, then, 59 is the 15th term in the sequence <Tex>(a_n)</tex>.

------------------

<Example type="example" num="4"/>

Find the number of terms in the arithmetic sequence: <Tex>1; \\; 4; \\; 7; \\; \ldots \ldots ;\\; 91</Tex>

<Example type="solution"/>

Here we have a finite arithmetic sequence, where the common difference d is 3, and the first item is <Tex>a_1=1</Tex>. We use the general term formula to calculate the number of terms in this sequence.

<Tex block>
\begin{align*}
a_n &= a_1+(n-1)d \\\
91 &= 1+(n-1)\times 3 \\\
91 &= 1+3n-3 \\\
91 &= 3n-2 \\\
3n &= 91+2=93 \\\
n &= \frac{93}{3} \\\
n &= 31
\end{align*}
</Tex>

So this sequence contains 31 terms.

-----------------

The general term formula enables us to calculate the value of nth term, if we reformulate it further, we get another formula that calculates the number of terms in a finite arithmetic sequence.

<Tex block>
\begin{align*}
a_n &= a_1+(n-1)d \\\
a_n-a_1 &= (n-1)d \\\
n-1 &= \frac{a_n-a_1}{d} \\\
n &= \frac{a_n-a_1}{d}+1
\end{align*}
</Tex>

<Note>

To determine the number of terms for a finite arithmetic sequence, we use the following formula:

<Tex block>n=\frac{a_n-a_1}{d}+1</Tex>

Whereas: <Tex>a_1</Tex> is the first term, <Tex>a_n</Tex> is the last term and <Tex>d</Tex> is the common difference.

</Note>

----------------

<Example type="example" num="5"/>

How many two-digit numbers are divisible by 3?

<Example type="solution"/>

The sequence of two-digit numbers divisible by 3 is a finite sequence. The first term of this sequence is 12 (the smallest two-digit number divisible by 3), and the last term of this sequence is 99 (the largest two-digit number divisible by 3).

The sequence is:

<Tex block>12; \\; 15; \\; 18; \\; ;\ldots ;\\; 96; \\; 99</Tex>

It's clear that is an arithmetic sequence with a common difference <Tex>d = 3</Tex>.

So that:

<Text block>
\begin{align*}
a_n &=a_1+(n-1)d \\\
99 &= 12+(n-1) \times 3 \\\
99 &= 12+3n-3 \\\
99 &= 3n+9 \\\
3n &= 90 \\\
n &= \frac{90}{3} \\\
n &= 30 
\end{align*}
</Tex>

Therefore, athe sequence consisting of 30 terms of two numbers are divisible by 3.

learn how to find general term of a sequence using arithmetic sequence formula. Various examples with solution that you can come across.

Arithmetic Sequence Formula: General Term

# Arithmetic Sequence Formula: Advanced General Term

<Related>
<Link to="../intermediate-algebra/arithmetic-sequence-definition-and-examples" as="/lessons/intermediate-algebra/arithmetic-sequence-definition-and-examples">Arithmetic Sequence (Progression): Definition and Examples</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-formula-general-term" as="/lessons/intermediate-algebra/arithmetic-sequence-formula-general-term">Arithmetic Sequence Formula: How to Find General Term</Link>
</Related>

In Arithmetic Sequences: General Term lesson, we saw that the general term formula is written as:

<Tex block>a_n=a_1+(n-1)d</Tex>


But what if we don't know the value of the first term <Tex>a_1</Tex>?

Can this formula be more generalized?

This is what we will answer in this lesson.

We know that: 

<Tex block>
\begin{equation}
a_n=a_1+(n-1)d \tag{1}
\end{equation}
</Tex>

If we take <Tex>p</Tex>, so that <Tex>p < n </Tex>. Similarly we get:

<Tex block>
\begin{equation}
a_p = a_1+(p-1)d \tag{2}
\end{equation}
</Tex>

Subtracting equations (1) and (2) side by side:

<Tex block>
\begin{align*}
a_n-a_p &= a_1+(n-1)d-a_1-(p-1)d \\\
a_n-a_p &= nd-d-pd+d \\\
a_n-a_p &= nd-pd \\\
a_n-a_p &=(n-p)d \\\
a_n &= a_p+(n-p)d
\end{align*}
</Tex>

This formula is also known as the general term of the sequence <Tex>(a_n)</Tex> where <Tex>p < n.</Tex>

<Note>
Let <Tex>(a_n)</Tex> an arithmetic sequence and <Tex>p</Tex> a natural number with <Tex>p<n.</Tex>
<Tex block>a_n = a_p+(n-p)d</Tex>

</Note>

-------------

<Example type="example" num="1"/>

Let <Tex>(a_n)</Tex> an arithmetic sequence in which <Tex>a_11=34</Tex> and <Tex>d=3</Tex>. Find <Tex>a_3</Tex>.

<Example type="solution"/>

In this example we only know the 11th term and d. What is required is to calculate the 3rd term.(Note carefully that the first term <Tex>a_1</Tex> is unknown).

To solve this problem we apply the above generalized general term property with <Tex>n=11</Tex> and <Tex>p=3</Tex>.

<Tex block>
\begin{align*}
a_n &= a_p+(n-p)d \\\
a_{11} &= a_3+(11-3)\times3 \\\
34 &= a_3+8\times3 \\\
34 &= a_3+24 \\\
a_3 &= 34-24 \\\
a_3 &= 10
\end{align*}
</Tex>

---------

<Example type="example" num="2"/>

Let <Tex>(a_n)</Tex> an arithmetic sequence in which <Tex>a_5=14</Tex> and <Tex>a_{10}=34</Tex>. Find <Tex>d</Tex>.

<Example type="solution"/>
To solve this problem we apply the above generalized general term property with <Tex>n=10</Tex> and <Tex>p=5</Tex>

<Tex block>
\begin{align*}
a_n &= a_p+(n-p)d \\\
a_{10} &= a_5+(10-5)\times d \\\
34 &= 14+5\times d \\\
5d &= 34-14 \\\
d &= \frac{20}{5} \\\
d &= 4
\end{align*}
</Tex>

---------

<Example type="example" num="3"/>

Let <Tex>(a_n)</Tex> an arithmetic sequence in which <Tex>a_9-a_2 = 42</Tex>. Find <Tex>a_{10}-a_7</Tex>.

<Example type="solution"/>
Pay attention in this example, what is required is the calculation of <Tex>a_{10}-a_7</Tex>, not the terms <Tex>a_{10}</Tex> and <Tex>a_7</Tex>

Let's find d first:

<Tex block>
\begin{align*}
a_n &= a_p+(n-p)d \\\
a_9 &= a_2+(9-2)d \\\
a_9-a_2 &= 7d \\\
42 &= 7d \\\
d &= \frac{42}{7} \\\
d &= 6
\end{align*}
</Tex>

Thus,

<Tex block>
\begin{align*}
a_{10} &=a_7+(10-7) \times6 \\\
a_{10}-a_7 &= 3\times6 \\\
a_{10}-a_7 &= 18
\end{align*}
</Tex>

---------

<Example type="example" num="4"/>

We consider the following arithmetic sequence: <Tex>4; \\; x; \\; y; \\; z; \\; 24</Tex>

Find <Tex>x, \\; y</Tex> and <Tex>z</Tex>

<Example type="solution"/>
Let <Tex>a_1 = 4</Tex> and <Tex>a_5 = 24</Tex>.

<Tex block>
\begin{align*}
a_5 &= a_1+(5 -1) \times d \\\
24 &= 4+4 \times d \\\
4d &= 20 \\\
d &= \frac{20}{4} \\\
d &= 5
\end{align*}
</Tex>

Thus,

<Tex block>
\begin{align*}
x &= 4+5 = 9 \\\
y &= 9+5 = 14 \\\
z &= 14+5 = 19
\end{align*}
</Tex>

---------

<Example type="example" num="5"/>

Let <Tex>(a_n)</Tex> an arethmitic sequence in wich <Tex>a_8=10</Tex>

Find <Tex>a_2</Tex><Tex>+a_{14}</Tex>,

<Example type="solution"/>

In this example, we have only one parameter, which is <Tex>a_8</Tex>.

What is required is to specify <Tex>a_2</Tex><Tex>+a_{14}</Tex> and not <Tex>a_2</Tex> and <Tex>a_{14}</Tex>.

Applying the formula. 

We write <Tex>a_2</Tex> in terms of <Tex>a_8</Tex>:

<Tex block>
\begin{align*}
a_2 &= a_8+(2-8)d \\\
a_2 &= a_8-6d \tag{3}
\end{align*}
</Tex>

And <Tex>a_{14}</Tex> in terms of <Tex>a_8</Tex>:

<Tex block>
\begin{align*}
a_{14} &= a_8+(14-8)d \\\
a_{14} &= a_8+6d \tag{4}
\end{align*}
</Tex>

Now, we adding the formula (1) and the formula (2) side by side:

<Tex block>
\begin{align*}
a_{14}+a_2 &= a_8-6d+a_8+6d \\\
&= 2 \times a_8 \\\
&= 2\times 10 \\\
&=20
\end{align*}
</Tex>

This lesson show you the generalized arithmetic sequence formula and how to use it in advanced problems.

Arithmetic Sequence Formula: Advanced General Term

# Arithmetic Sequence Formula: Arithmetic Mean

<Related>
<Link to="../intermediate-algebra/arithmetic-sequence-definition-and-examples" as="/lessons/intermediate-algebra/arithmetic-sequence-definition-and-examples">Arithmetic Sequence (Progression): Definition and Examples</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-formula-general-term" as="/lessons/intermediate-algebra/arithmetic-sequence-formula-general-term">Arithmetic Sequence Formula: How to Find General Term</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-formula-advanced-general-term" as="/lessons/intermediate-algebra/arithmetic-sequence-formula-advanced-general-term">Arithmetic Sequence Formula: Advanced General Term</Link>
</Related>

We consider the following <Tex>(a_n)</Tex> arithmetic sequence:

<Tex block>...; \\; {a}\_{p-2}; \\; a\_{p-1}; \\; a\_{p}; \\; a\_{p+1}; \\; a\_{p+2}; \\; ...</Tex>

It is clear that <Tex>a_{p+k}</Tex> and <Tex>a_{p-k}</Tex> are two terms that are located at the same distance from <Tex>a_p</Tex>. (<Tex>a_{p+k}</Tex> on the right and <Tex>a_{p-k}</Tex> on the left).

Let's write <Tex>a_{p+k}</Tex> using the general term formula:

<Tex block>a_{p+k}=a_p+kd \tag{1}</Tex>

And let's write <Tex>a_{p-k}</Tex> using the general term formula:

<Tex block>a_{p-k}=a_p-kd \tag{2}</Tex>

Now, we add the formulas (1) and (2) side by side:

<Tex block>
\begin{align*}
a\_\{p+k}+a\_\{p-k} &=a_p+kd+a_p-kd \\\
a\_\{p+k}+a\_\{p-k} &= 2a_p
\end{align*}
</Tex>

And consequently:

<Tex block>
a\_p = \frac{a\_{p+k}+a\_{p-k}}{2}
</Tex>

This formula is called the **mean value of an arithmetic sequence**.

This means that the term <Tex>a_p</Tex> is located exactly in the middle between the two terms <Tex>a_{p+k}</Tex> and <Tex>a_{p-k}</Tex>.

<Note>
The mean value of an arithmetic sequence is:
<Tex block>a\_p = \frac{a\_{p+k}+a\_{p-k}}{2}</Tex>

</Note>

As a result of the above, the sequence <Tex>\\{ a,\\; b, \\; c \\}</Tex> is an arithmetic sequence if and only if <Tex>b=\frac{a+c}2</Tex>

-----------

Here are some examples:

<Example type="example" num="1"/>

Let <Tex>\\{ 5,\\; x, \\; 19 \\}</Tex> three consecutive numbers in an arithmetic sequence. Let's calculate <Tex>x</Tex>.

<Example type="solution"/>

we've got: <Tex>a_1=5 \\; ; \\; a_2=x \\; ; \\; a_3=19</Tex>

Using the mean arithmetic formula:<Tex>x=a_2=\frac{a_1+a_3}{2}=\frac{5+19}{2}=12</Tex>

-----------

<Example type="example" num="2"/>

Find the general term <Tex>(a_n)</Tex> of an arithmetic sequence such that <Tex>a\_5+a\_{21}=106</Tex> and <Tex>a_9=37</Tex>.

<Example type="solution"/>

The term 13 is located on the same distance between the terms <Tex>a_5</Tex> and <Tex>a_{21}</Tex>, so:

<Tex block>a\_{13}=\frac{a\_5+a\_{21}}{2}=\frac{106}{2}=53</Tex>

Let's calculate <Tex>d</Tex>

<Tex block>
\begin{align*}
a_n &=a_p+(n-p)d \\\
a_{13} &= a_9+(13-9)d \\\
53 &= 37 +4d \\\
4d &= 53-37=16 \\\
d &= 4
\end{align*}
</Tex>

Then, the general term is:

<Tex block>
\begin{align*}
a_n &=a_p+(n-p)d \\\
a_n &= a_9+(n-9) \times 4 \\\
a_n &= 37 +4n-36 \\\
a_n &=4n+1
\end{align*}
</Tex>

this lesson show you how to find the arithmetic mean value of an arithmetic sequence using formula of mean.

Arithmetic Sequence Formula: Arithmetic Mean Formula

# Exponent Notation: Definition and Rules

In this lesson, you will learn about the exponent notation, its components and how to calculate it, as well as extract the most important properties of powers used to simplify algebraic expressions.
Power, quite simply, is multiplying a number of times by itself. For example, we take the power <Tex>5^3</Tex> as the product result of <Tex>5\times 5 \times 5</Tex>, where the small, raised number <Tex>3</Tex> to the right of the base number indicates the number of times the base number is multiplied by itself.

<Image src="/imgs/lessons/intermalgebra/exponent-notation.png" alt="exponent notation rules" width='900px' height='400px' />

<Note>
Let <Tex>x</Tex> a real number and <Tex>n</Tex> a whole positif number, then the nth power of <Tex>x</Tex> is:

<Tex block>a^n=\underbrace{a \times a \times a \times \ldots \times a}\_{\text{n factors}}</Tex>

The number <Tex>x</Tex> is called the **base**, and <Tex>n</Tex> is called the **exponent**.

</Note>

-------------------

<Example type="example" num="1"/>

Let's calculate the following powers:

* <Tex>5^3=5\times 5\times 5 = 125</Tex>

* <Tex>2^5=2\times 2\times 2 \times 2\times 2=32</Tex>

* <Tex>\left( \frac{1}{4} \right)^4=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} =\frac{1}{256}</Tex>

Now lets do the inverse, transform the follwing numbers to exponnent notation:

* <Tex>27=3\times 3\times 3=3^3</Tex>

* <Tex>-216=(-6)\times (-6)\times (-6)=(-6)^3</Tex>

## Exponential Notation Rules

### The Product Rule

From the above definition and examples, we can easily calculate the product of the two powers. we've got:

<Tex>a^2 \times a^4=(a\times a\times a\times a)\times (a\times a\times a\times a\times a\times a)=a^6</Tex>

How do you think we got the power that the exponent is <Tex>6</Tex>?

You will no doubt have noticed that if we add the exponents of the powers <Tex>a^2</Tex> and <Tex>a^4</Tex>, we get 6. From this example, we can deduce the following property:

<Note>
Let <Tex>a</Tex> a real number and <Tex>m</Tex> and <Tex>n</Tex> are integers. We have:

<Tex block>a^m \times a^n=a^{m+n}</Tex>

When multiplying exponential expressions with the same base, add the exponents. Use this sum as the exponent of the common base.
</Note>

<Example type="example" num="2"/>

Multiply each expression using the product rule:

* <Tex>3^2\times 3^3</Tex>
* <Tex>(4x^5y^7)\centerdot (3x^2y^3z^4)</Tex>

<Example type="solution"/>

* <Tex block>3^2 \times 3^3=3^{2+3}=3^5=243</Tex>
* <Tex block>
\begin{align*}
(4x^5y^7)\centerdot (3x^2y^3z^4)&=4\times x^5\times y^7\times 3\times x^2\times y^3\times z^4 \\\
&=4\times 3\times x^5\times x^2\times y^7\times y^3\times z^4 \\\
&=12\times x^{5+2}\times y^{7+3}\times z^4 \\\
&=12x^{7}y^{10}z^4
\end{align*}
</Tex>

-------------------

### The Quotient Rule

From the above definition and examples, we can easily calculate the quotient of the two powers. we've got:

<Tex>\frac{a^5}{a^2}=\frac{a\times a\times a\times a\times a}{a\times a}=a\times a\times a=a^3</Tex>

How do you think we got the power that the exponent is <Tex>3</Tex>?

In this example, we have subtracted the exponent of the power in the denominator from the exponent of the power in the numerator (<Tex>5-2=3</Tex>). From this example, we can deduce the following property:

<Note>
Let <Tex>a</Tex> a non-zero real number and <Tex>m</Tex> and <Tex>n</Tex> are integers. We have:

<Tex block>\frac{a^m}{a^n}=a^{m-n}</Tex>

When dividing exponential expressions with the same nonzero base, subtract the exponent in the denominator from the exponent in the numerator. Use this difference as the exponent of the common base.

</Note>

<Example type="example" num="3"/>

Divide each expression using the quotient rule:

* <Tex>\frac{8^3}{8^7}</Tex> 
 

* <Tex>\frac{(-3)^{10}}{(-3)^{-6}}</Tex> 
 

* <Tex>\frac{x^8y^7z^2}{x^3y^8}</Tex>

<Example type="solution"/>

* <Tex block>\frac{8^3}{8^7}=8^{3-7}=8^{-4}</Tex>
* <Tex block>\frac{(-3)^{10}}{(-3)^{-6}}=(-3)^{10-(-6)}=(-3)^{16}</Tex>
* <Tex block>
\begin{align*}
\frac{x^8y^7z^2}{x^3y^8}&=x^{8-3}y^{7-8}z^2 \\\
&=x^5y^{-1}z^2
\end{align*}
</Tex>

-------------------

### The Products-to-Powers Rule for Exponents

In this section, the rule of exponents applies when we are raising a product to a power. Let's take the following example:

The base of the power <Tex>(3a)^4</Tex> is the product <Tex>3a</Tex>.

applying the formula in the definition:

<Tex block>
\begin{align*}
(3a)^4&=3a\times 3a\times 3a\times 3a \\\
&=3\times 3\times 3\times 3\times a\times a\times a\times a \\\
&=3^4a^4
\end{align*}
</Tex>

Now, how did we get the answer <Tex>3^4a^4</Tex>?

If you notice, we've removed the parentheses and raised both factors,<Tex>3</Tex> and <Tex>a</Tex>, to the fourth power. From this example, we can deduce the following property:

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> are real numbers and <Tex>n</Tex> an integer. We have:

<Tex block>(ab)^n=a^nb^n</Tex>

When a product is raised to a power, raise each factor to that power.

</Note>

<Example type="example" num="4"/>

* Simplify: <Tex>(2x)^5</Tex>
* Simplify: <Tex>(-3xy)^3</Tex>

<Example type="solution"/>

* <Tex>(2x)^5=2^5x^5=32x^5</Tex>
* <Tex>(-3xy)^3=(-3)^3x^3y^3=-27x^3y^3</Tex>

-------------------

### The Quotients-to-Powers Rule for Exponents

In this section, the rule of exponents applies when we are raising a quotient to a power. Let's take the following example:

The base of the power <Tex>\left( \frac{3}{a} \right)^4</Tex> is the quotient <Tex>\frac{3}{a}</Tex>.

applying the formula in the definition:

<Tex block>
\begin{align*}
\left( \frac{3}{a} \right)^4&=\frac{3}{a} \times \frac{3}{a} \times \frac{3}{a} \times \frac{3}{a} \\\
&=\frac{3\times 3\times 3\times 3}{a\times a\times a\times a} \\\
&=\frac{3^4}{a^4}
\end{align*}
</Tex>

Now, how did we get the answer <Tex>\frac{3^4}{a^4}</Tex>?

If you notice, we've removed the parentheses and raised denominator and numerator,<Tex>3</Tex> and <Tex>a</Tex>, to the fourth power. From this example, we can deduce the following property:

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> are real numbers with <Tex>b\neq 0</Tex> and <Tex>n</Tex> an integer. We have:

<Tex block>\left( \frac{a}{b} \right)^n=\frac{a^n}{b^n}</Tex>

When a quotient is raised to a power, raise the numerator to that power and divide by the denominator to that power.

</Note>

<Example type="example" num="5"/>

* Simplify: <Tex>\left( \frac{x}{-4} \right)^4</Tex>
* Simplify: <Tex>\left( \frac{9x}{y} \right)^2</Tex>

<Example type="solution"/>

* <Tex>\left( \frac{x}{-4} \right)^4=\frac{x^4}{(-4)^4}=\frac{x^4}{256}</Tex> 
 

* <Tex>\left( \frac{9x}{y} \right)^2=\frac{9^2x^2}{y^2}=\frac{81x^2}{y^2}</Tex>

-------------------

### The Negative-Exponent Rule

In the definition of the exponontial notation, We saw that the exponent is a positive integer, but what about if you see a
negative exponent? We can easily calculate a power with negative exponent follwing the below example:
We simplify the quotient <Tex>\frac{5^3}{5^5} with two different ways.

<Tex block>\frac{5^3}{5^5}=\frac{5\times 5\times5}{5\times 5\times 5\times 5\times 5}=\frac{1}{5^2}</Tex>

<Tex block>\frac{5^3}{5^5}=5^{3-5}=5^{-2}</Tex>

This means that <Tex>\frac{5^3}{5^5}=5^{-2}=\frac{1}{5^2}</Tex>. A negative exponent in the numerator becomes a positive
exponent when moved to the denominator and vice versa.

<Note>

Let <Tex>a</Tex> be a nonzero real number and <Tex>n</Tex> be any integer.

<Text block>a^{-n}=\frac{1}{a^n} \quad and \quad a^{n}=\frac{1}{a^{-n}}</Tex>

</Note>

<Example type="example" num="6"/>

Write each expression without negative exponents, and evaluate if possible

* <Tex>4^{-2}</Tex>
* <Tex>\frac{1}{2^{-5}}</Tex>
* <Tex>\left( \frac{a}{b} \right)^{-3}</Tex>
* <Tex>(xy)^{-1}</Tex>
* <Tex>xy^{-1}</Tex>

<Example type="solution"/>

* <Tex>4^{-2}=\frac{1}{4^2}=\frac{1}{16}</Tex> 
 
* <Tex>\frac{1}{2^{-5}}=2^5=32</Tex>
* <Tex>\left( \frac{a}{b} \right)^{-3}=\frac{a^{-3}}{b^{-3}}=\frac{b^{3}}{a^{3}}=\left( \frac{b}{a} \right)^{3}</Tex>
* <Tex>(xy)^{-1}=\frac{1}{xy}</Tex>
* <Tex>xy^{-1}=x\times \frac{1}{y}=\frac{x}{y}</Tex>

-------------------

### The Zero-Exponent Rule

There is a special case in calculating powers, when the exponent is zero. How can we calculate a power taht zero as an exponent? As in the previous paragraphs, we start from the following two examples:

<Tex block>x^4x^{-4}=\frac{x^4}{x^4}=1</Tex>
<Tex block>x^4x^{-4}=x^{4-4}=x^0</Tex>

This means that <Tex>x^0=1</Tex>

<Note>

Let <Tex>a</Tex> a nonzero real number.

<Tex block>a^0=1</tex>

</Note>

<Example type="example" num="7"/>

Here are examples involving simplification using the zero-exponent rule:

<Tex>100^0=1 \\; ; \\; (-100)^0=1 \\; ; \\; -100^0=-1 \\; ; \\; \left( \frac{15}{x} \right)^0=1</Tex>

-------------------

### Things to remember about exponents

There are some special cases which are very important in simplifying algebraic expressions. We can prove these special cases in the same way in the previous proofs.

<Note>

1. Any base without an exponent has an “invisible” exponent of 1.
<Tex block>x^1=x</Tex>

2. Any power with a base 1, equals 1.
<Tex block>1^n=1</Tex>

3. if the exponent is an even number:
<Tex block>(-x)^n=x^n</Tex>

4. if the exponent is an odd number:
<Tex block>(-x)^n=-x^n</Tex>

</Note>

Learn exponential notation, exponent rules and properties with some examples with solution.

Exponent Notation: Definition and Rules

# The polynomials: Definition, Types, Degree and Properties

<Related>
<Link to="../basic-algebra/properties-of-addition-and-multiplication" as="/lessons/basic-algebra/properties-of-addition-and-multiplication">Properties of Addition and Multiplication: Commutative, Associative and Identity</Link> 
<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 
<Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Simplifying algebraic expressions by Combining Like Terms</Link> 
<Link to="../intermediate-algebra/exponent-rules-properties" as="/lessons/intermediate-algebra/exponent-rules-properties">Exponent Notation: Definition and Rules</Link>
</Related>

If there is one thing that you should pay close attention to during your journey to learn the basics of mathematics, it will undoubtedly have to do with **polynomials** due to their important role in dealing with different algebraic expressions. Polynomials are included in many mathematical concepts.

In this lesson, we will try to discuss about the definition of the concept of polynomials and its standard form, and we will explain the types of polynomials and the concept of "the degree of polynomial". Finally, we will see some of the properties of the polynomials.

## Polynomials and its standard form

<Note>

A polynomial is the sum of multiple terms and each term is a mononomial.

</Note>

Based on this definition, we must first get to know the concept of term in detail because it is the basic building block of which polynomials are composed.

<Note>

The term is a product of a number (which is the coefficient) and one or more variables raised to powers. The exponent is a natural integer.

</Note>

Take, for example:

* The term <Tex>x^2</Tex> has coefficient 1.

* The term <Tex>-3x^4</Tex> has coefficient 4.

* The term <Tex>-x^3y</Tex> has coefficient -1.

**Like terms** in an algebraic expression are terms that consist of the same variable or variables and are raised to the same power.

Take, for example:

* Like terms: <Tex>-9x^2 \\; ; \\; 5x^2 \\; ; \\; x^2.</Tex>

* Unlike terms: <Tex>-9x^4 \\; ; \\; 5x^3 \\; ; \\; x^2.</Tex>

* Unlike terms: <Tex>x^2 \\; ; \\; y^2 \\; ; \\; z^2.</Tex>

By defining the term, we can now say that a polynomial is the sum of a finite number of terms.

Take, for example:

* <Tex>-3x^4+5x^3-x^2+\frac{x}{2}-1</Tex> is a polynomial in <Tex>x</Tex>.

* <Tex>2x^2y+3xy^3-xy+x</Tex> is a polynomial in <Tex>x</Tex> and <Tex>y</Tex>.

**N.B:** The terms of a polynomial cannot have variables in a denominator.

Take, for example: 

* <Tex>x^3+2x^2+3x-\frac{6}{x}</Tex> is not a polynomial.

## Types of polynomials

There are three types to distinguish between types of polynomials.

**Monomials:** It is an algebraic expression consisting of only one term.

For example, <Tex>x^2 \\; ; \\; 13x^3y \\; ; \\; -xyz^2</Tex>

**Binomials:** It is an algebraic expression consisting of only two terms.

For example, <Tex>x^2+1 \\; ; \\; -5x^5+x^3 \\; ; \\; -x^2y-xy^2</Tex>

**Trinomials:** It is an algebraic expression consisting of only three terms.

For example, <Tex>x^2+2x+1 \\; ; \\; x^2+y^2+xy</Tex>

Binomials and trinomials can also be called polynomials.

## Degree of a polynomial

To return to the concept of the term, in the first paragraph above, we said that the term is a number multiplied by a variable or several variables raised to an exponent. So what is the degree of a term?

The degree of the term is the sum of the exponents of the variable(s) that exist in this term.

Take, for example: 

* The degree of <Tex>-3x^4 y^3 z^5</Tex> is <Tex>4+3+5=12</Tex>.
 

* The degree of <Tex>5xy^2 z^7</Tex> is <Tex>1+2+7=10</Tex>.


* The degree of <Tex>2x</Tex> is <Tex>1</Tex>.

* The degree of <Tex>2</Tex> is <Tex>0</Tex>.

* More examples

<Image src="/imgs/lessons/intermalgebra/degree_term.png" alt="degree of a one term" width={650} height={448} />

So, what is the degree of a polynomial?

<Note>

Degree of a polynomial is the greatest degree of any terms of the polynomial.

</Note>

Take, for example: 

* The degree of <Tex>5x^8 -3x^4 +2x^3</Tex> is <Tex>8</Tex>.

* The degree of <Tex>3x^2 y^3 -5x^6 y +4xy^9</Tex> is <Tex>1+9=10</Tex>.

* More examples

<Image src="/imgs/lessons/intermalgebra/polynomial_degree.png" alt="degree of a polynomial" width={500} height={349} />


<Image src="/imgs/lessons/intermalgebra/polynomial_degree_2.png" alt="degree of a polynomial" width={500} height={349} />

## Properties of a polynomials

Now that we know the definition of polynomials and how to determine their degree, we will now move on to some properties about adding, subtracting and multiplying polynomials.
In this summary, we will try to take some examples of polynomials with only one unknown in order to simplify the idea of performing the three operations mentioned above.

### Adding and subtracting of a two polynomials

<Note>

The sum or subtraction of two <Tex>P</Tex> polynomials and <Tex>Q</Tex> is a polynomial <Tex>R</Tex>.

And write <Tex>R=P+Q</Tex> or <Tex>R=P-Q</Tex>

The degree of the polynomial <Tex>R</Tex> is the largest degree of <Tex>P</Tex> and <Tex>Q</Tex>.

And write <Tex>deg(R)=max(deg(P), deg(Q))</Tex>

</Note>

<Example type="example" num="1"/>

Let <Tex>P</Tex> and <Tex>Q</Tex> are two polynomials in <Tex>x</Tex>, so that:

<Tex block>P=6x^3+4x^2-x</Tex>

<Tex block>Q=-3x^2+7x-1</Tex>

<Alert type="infos"> 

Reminder:

- If the parentheses are preceded by a (+) we remove the parentheses without changing anything.

- If the parentheses are preceded by (- ), we remove the parentheses and then change the signs in those parentheses to their opposite.

- We simplify algebraic expressions using the properties of **Combining Like Terms** and **Distributivity**.

</Alert>

**Let's Adding P and Q:**

<Text block>
\begin{align*}
R&=P+Q \\\
&=(6x^3+4x^2-x)+(-3x^2+7x-1) \\\
&=6x^3+4x^2-x-3x^2+7x-1 \\\
&=6x^3+(4-3)x^2+(7-1)x-1 \\\
&=6x^3+x^2+6x-1
\end{align*}
</Tex>

**Let's Subtracting P and Q:**

<Text block>
\begin{align*}
R&=P-Q \\\
&=(6x^3+4x^2-x)-(-3x^2+7x-1) \\\
&=6x^3+4x^2-x+3x^2-7x+1 \\\
&=6x^3+(4+3)x^2+(-1-7)x+1 \\\
&=6x^3+7x^2-8x+1
\end{align*}
</Tex>

So as you can see in the two operations, <Tex>R</Tex> is also a polynomial and its degree is 3 (<Tex>P</Tex> has degree 3 and <Tex>Q</Tex> has degree 2, so, <Tex>R</Tex> has degree 3).

**Let's Multiplying P and Q:**

<Text block>
\begin{align*}
R&=P \times Q \\\
&=(6x^3+4x^2-x)(-3x^2+7x-1) \\\
&=6x^3 \times (-3x^2)+6x^3 \times 7x+6x^3 \times (-1)+4x^2 \times (-3x^2)+4x^2 \times 7x+4x^2 \times (-1)-x \times (-3x^2)-x \times 7x-x \times (-1) \\\
&=-18x^5+42x^4-6x^3-12x^4+42x^3-4x^2+3x^3-7x^2+x \\\
&=-18x^5+(42-12)x^4+(-6+42+3)x^3+(-4-7)x^2+x \\\
&=-18x^5+30x^4+39x^3-13x^2+x
\end{align*}
</Tex>

So as you can see, <Tex>R</Tex> is also a polynomial and its degree is 5 (<Tex>P</Tex> has degree 3 and <Tex>Q</Tex> has degree 2, so, <Tex>R</Tex> has degree 3+2=5).

<Alert type="infos>

For the division, we will try to explain it in another lesson because it is not quite as simple as the above-mentioned operations.

</Alert>

You will learn the basics of plynomials and its standard form, the types (monomials, binomials and trinomials) the degree of polynomial and some properties

Polynomials: Definition, Types, Degree and Properties

# Factoring Polynomials Out the Greatest Common Factor

<Image src="/imgs/lessons/intermalgebra/factoring.jpeg" alt="factoring polynomials - mathsweet.com" objectFit="fill" width="600" height="530" placeholder="blur" 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This chapter describes how to factor algebraic expressions with the greatest common factor (gcf). There are several different ways to examine in detail one at a time, depending on the expression we want to factor. The exercises available in this chapter will allow you to apply each of the methods studied.

<Note>

Factoring an expression is to transform a sum or a difference into a product.

</Note>

By this definition, the algebraic expression must consist of at least two terms in order to find the common factor between these terms.

We take a simple examples of algebraic expressions with the number of their terms:

- In <Tex>8x+18</Tex>, we have two terms: 18 and <Tex>8x</Tex>.

- In <Tex>6(x+1)^2-3</Tex>, we have two terms: <Tex>6(x+1)^2</Tex> and <Tex>3</Tex>.

- In <Tex>5x^2+10x-25</Tex>, we have three terms: <Tex>5x^2</Tex>, <Tex>10x</Tex> and <Tex>25</Tex>.

<Alert>

Factoring is useful in several cases, including solving equations and studying functions.

</Alert>

## Factoring with a common factor: esay case

In this lesson we will need the greatest common factor, so I invite you to go to Lesson <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">hwo to find gcf</Link> before you read on here.

<Note>

the common factor It is quite simply a number, a letter or an expression (with or without parentheses) that is found in each of the terms.

</Note>

taking simple examples to fully understand:

- In <Tex>5x+5y</Tex>, the common factor is 5.

- In <Tex>13x+7x</Tex>, the common factor is <Tex>x</Tex>.

- In <Tex>a \times k + b \times k</Tex>, the common factor is <Tex>k</Tex>.

- In <Tex>7x(4x+1)-y(4x+1)</Tex>, the common factor is <Tex>(4x+1)</Tex>.


<Note>

To factorize by a common factor, we use the reciprocal of the property of the distributivity.

<Tex block>k \times a+k \times b=k \times (a+b)</Tex>

</Note>

Now we know what the common factor is and how to define it in an algebraic expression. Now let's work out simple algebraic expressions where the common factor is a simple number or variable.

<Example type="example" num="1"/>

Factor the following expressions:

* <Tex>7x+7y</Tex>

* <Tex>7x+7</Tex>

* <Tex>2x-2y</Tex>

* <Tex>9x+x \sqrt{2}</Tex>

* <Tex>a^2b+a^2c</Tex>

* <Tex>\frac{x}{3}+\frac{y}{3}</Tex>

<Example type="solution"/>

* In <Tex>7x+7y</Tex>, the common factor is 7, so:

<Tex block>
\begin{align*}
7x+7y&=7\times x+7\times y \\\
&=7\times (x+y)
\end{align*}
</Tex>

* In <Tex>7x+7</Tex>, the common factor is 7, so:

<Tex block>
\begin{align*}
7x+7&=7\times x+7\times 1 \\\
&=7\times (x+1)
\end{align*}
</Tex>

* In <Tex>2x-2y</Tex>, the common factor is 2, so:

<Tex block>
\begin{align*}
2x-2y&=2\times x-2\times y \\\
&=2\times (x-y)
\end{align*}
</Tex>

* In <Tex>9x+x \sqrt{2}</Tex>, the common factor is <Tex>x</Tex>, so:

<Tex block>
\begin{align*}
9x+x \sqrt{2}&=9\times x+\sqrt{2}\times x \\\
&=x\times (9+\sqrt{2})
\end{align*}
</Tex>

* In <Tex>a^2b+a^2c</Tex>, the common factor is <Tex>a^2</Tex>, so:

<Tex block>
\begin{align*}
a^2b+a^2c&=a^2\times b+a^2\times c \\\
&=a^2\times (b+c)
\end{align*}
</Tex>

* In <Tex>\frac{x}{3}+\frac{y}{3}</Tex>, the common factor is <Tex>\frac{1}{3}</Tex>, so:

<Tex block>
\begin{align*}
\frac{x}{3}+\frac{y}{3}&=\frac{1}{3} \times x +\frac{1}{3} \times y \\\
&=\frac{1}{3} \times (x+y)
\end{align*}
</Tex>

As you can see, the factor is put in front, then in the parenthesis we rewrite the starting expression without the common factor.

To check, we can take the expression on the right and expand it using the distributive property, we should find the expression on the left:

<Tex block>7\times(x+y)=7\times x+7\times y=7(x+y)</Tex>

## Factoring with a not obvious greatest common factor

there are some cases where the common factor is not obvious, it has to be brought out.

For example, the expression <Tex>8x+12y</Tex> we don't see the common factor, but we can make it appear by decomposing the terms.

Therefore, the purpose of this paragraph is to rewrite the expression in a different way in order to show the common factor.

Let's back to the privious expression, <Tex>8x+12y</Tex> can be rewritten in two different ways:

<Tex>
8x+12y=2\times 4x+2\times 6y\\\
8x+12y=4\times 2x+4\times 3y
</Tex>

In the fisrt line, the common factor is 2, in the second line, the common factor is 4. In this case, always, we choose the greatest common factor, that's mean we factor by 4.

<Tex>
8x+12y=4\times 2x+4\times 3y=4(2x+3y)
</Tex>

Note that there are common factors may be a bit more complicated to find it.

<Example type="example" num="2"/>

Factor the following expressions:

* <Tex>20x-5</Tex>

* <Tex>4xy+6xz</Tex>

* <Tex>28x^2+21x</Tex>

* <Tex>y \sqrt{x}+z\sqrt{x}</Tex>

* <Tex>-x-1</Tex>

<Example type="solution"/>

* Its obvious that <Tex>gcf(20;5)=5</Tex>, therefore:

<Tex block>
\begin{align*}
20x-5&=5\times4x-5\times1 \\\
&=5(4x-1)
\end{align*}
</Tex>

* Its obvious that gcf(4;6)=2, and <Tex>x</Tex> is another common factor, therefore, the greatest common factor is <Tex>2x</Tex>:

<Tex block>
\begin{align*}
4xy+6xz&=2\times 2\times x \times y+2\times 3\times x \times z \\\
&=2x(2y+3z)
\end{align*}
</Tex>

* In <Tex>28x^2+21x</Tex>, its clear that <Tex>gcf(28;21)=7</Tex>, and <Tex>x</Tex> is another common factor, therefore, the greatest common factor is <Tex>7x</Tex>:

<Tex block>
\begin{align*}
28x^2+21x&=7\times4\times x\times x-7\times3\times x \\\
&=7x(4x+3)
\end{align*}
</Tex>

* In <Tex>y \sqrt{x}+z\sqrt{x}</Tex>, simply, the common factor is <Tex>sqrt{x}</Tex>, therefore:

<Tex block>
\begin{align*}
y \sqrt{x}+z\sqrt{x}&=\sqrt{x}\times y+\sqrt{x}\times z \\\
&=\sqrt{x}(y+z)
\end{align*}
</Tex>

* What do you think about this <Tex>-x-1</Tex> expression? Can you find the common factor? :

<Tex block>
\begin{align*}
-x-1&=(-1)\times x+(-1)\times1 \\\
&=-1\times (x+1) \\\
&=-(x+1)
\end{align*}
</Tex>

## Factoring polynomials by grouping

In this section, we will try to factor polynomials that containes parenthesis or brackets and the common factor will be the expression between the parenthesis. We apply the reciprocal property in the same way as above, but this time we call it **factoring by grouping**.

<Note>

To factorize by the common factor <Tex>(c+d)</Tex>, we use the reciprocal of the property of the distributivity.

<Tex block>a \times (c+d)+b \times (c+d)=(a+b) \times (c+d)</Tex>

</Note>

For example, look closely at this expression: <Tex>2(x+1)+y(x+1)</Tex>. You've no doubt noticed that the common factor is <Tex>(x+1)</Tex>. So,

<Tex block>2(x+1)+y(x+1)=(x+1)(2+y)</Tex>

<Example type="example" num="3"/>

Factor the following expressions:

* <Tex>6x(2x-1)+5y(2x-1)</Tex>

* <Tex>-7x(x+5)+(10x+1)(x+5)</Tex>

* <Tex>(4x-1)(y+1)+(-3x+5)(y+1)</Tex>

* <Tex>x(x^2+1)+y(x^2+1)-z(x^2+1)</Tex>

* <Tex>4x(\sqrt{x}+1)-2y(\sqrt{x}+1)</Tex>

<Example type="solution"/>

<Alert type="warning">

You may be need to be familiar with **adding and subtracting the polynomials**. I suggest you read this <Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">lesson</Link>.

</Alert>

* The common factor is <Tex>(2x-1)</Tex>:

<Tex block>
\begin{align*}
6x(2x-1)+5y(2x-1)&=(2x-1)(6x+5y)
\end{align*}
</Tex>

* The common factor is <Tex>(x+5)</Tex>:

<Tex block>
\begin{align*}
-7x(x+5)+(10x+1)(x+5)&=(x+5)[-7x+(10x+1)] \\\
&=(x+5)[-7x+10x+1] \\\
&=(x+5)(3x+5)
\end{align*}
</Tex>

* The common factor is <Tex>(y+1)</Tex>:

<Tex block>
\begin{align*}
(4x-1)(y+1)+(-3x+5)(y+1)&=(y+1)[(4x-1)+(-3x+5)] \\\
&=(y+1)[4x-1-3x+5] \\\
&=(y+1)(x+4)
\end{align*}
</Tex>

* The common factor is <Tex>(x^2+1)</Tex>:

<Tex block>
\begin{align*}
x(x^2+1)+y(x^2+1)-z(x^2+1)&=(x^2+1)(x+y+z)
\end{align*}
</Tex>

* The common factor is <Tex>(\sqrt{x}+1)</Tex>, right? Oh no!! look good at the expression, we have <Text>gcf(4;2)=2</Tex>, so, the common factor is <Tex>2(\sqrt{x}+1)</Tex>:

<Tex block>
\begin{align*}
4x(\sqrt{x}+1)-2y(\sqrt{x}+1)&=2\times 2\times(\sqrt{x}+1)-2\times y\times (\sqrt{x}+1) \\\
&=2(\sqrt{x}+1)(2-y)
\end{align*}
</Tex>

You will learn how to factoring polynomials out the greatest common factor (gcf)

Factoring Polynomials Out the Greatest Common Factor

# Factoring a Simple Quadratic Trinomials: <Tex>x^2+bx+c</Tex>

In the polynomial lesson, we saw that a trinomial is a special case of a polynomial, and a trinomial means an **algebraic expression of three terms**. In another lesson, we saw [how to factor polynomials using the common factor](/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor) (out of the gcd of the polynomial terms).

In this lesson, we will learn about how to factor a quadratic trinomial.

<Note>

A quadratic trinomial is every algebraic expression written as:

<Tex block>ax^2+bx+c</Tex>

</Note>

From the title of the lesson, you noticed that we called it “simple quadratic trinomial”, because we will take <Tex>a=1</Tex> in the trinomial expression as a special case and we get <Tex>x^2+bx+c</Tex>. The factoring will be very easy in this case. Just read the lesson to the end and re-do the examples.

<Alert>

In another lesson, we will explain how to factor a quadratic trinomial <Tex>ax^2+bx+c</Tex> with <Tex>a\neq1</Tex>.

</Alert>

## General Formula Of Factoring Quadratic Trinomial

<Note>

Let <Tex>a</Tex>, <Tex>b</Tex> and <Tex>x</Tex> be real numbers. we've got:

<Tex block>x^2+bx+c=(x+m)(x+n)</Tex>

</Note>

To factor the quadratic trinomial, we need to find the numbers <Tex>m</Tex> and <Tex>n</Tex>. Below we will see how to find them.

First, let's expand the expression by FOIL method <Tex>(x+m)(x+n)</Tex>:

<Tex block>
\begin{align*}
(x+m)(x+n)&=x^2+nx+mx+m.n \\\
&=x^2+(m+n)x+m.n \\\
&=x^2+bx+c
\end{align*}
</Tex>

After expanding and simplifying the expression, we can conclude that the number <Tex>b</Tex> is equal to <Tex>(m+n)</Tex>, and the number <Tex>c</Tex> is equal to the number <Tex>n \times m</Tex>. Thus, we get a simple system:

<Tex block>
\begin{cases}
 m+n=b \\\
 m \times n=c
\end{cases}
</Tex>

## How To Find n And m?

**Step1:** The equation <Tex>m \times n=c</Tex> gives us two important ideas:

- The first idea (The sign of the two numbers <Tex>m</Tex> and <Tex>n</Tex>)

 * If the number <Tex>c</Tex> is positive, then the two numbers <Tex>m</Tex> and <Tex>n</Tex> have the same sign.

 * If the number <Tex>c</Tex> is negative, then the two numbers <Tex>m</Tex> and <Tex>n</Tex> have different signs.

- The second idea (decomposition <Tex>c</Tex> into product of two numbers)

We do all the possible decompositions of <Tex>c</Tex> into product of two numbers. Having defined the sign previously, we can now only take the appropriate decompositions.

 * If <Tex>c</Tex> is positive, we take all of the decompositions that they factors out have the same sign.

 * If <Tex>c</Tex> is negative, we take all of the decompositions that they factors out have two different signs.

**Step2:** The equation <Tex>m + n=b</Tex> will enable us to determine the final decomposition from among the decompositions that we obtained in the first step, and from it we will get the two numbers m and n.

<Alert>

Note that if <Tex>a=-1</Tex>, we factor the quadratic trinomial by -1, then we apply the previous steps (See Example 5).

</Alert>

<Example type="example" num="1"/>

Let's factoring the quadratic trinomial: <Tex>x^2+8x+15</Tex>

<Example type="solution"/>

<Tex>x^2+8x+15=(x+m)(x+n)</Tex>

**Step1:** We decompose the number 15:

<Tex block>
\begin{align*}
15&=1\times15 \\\
15&=3\times5 \\\
15&=(-1)\times(-15) \\\
15&=(-3)\times(-5)
\end{align*}
</Tex>

**Step2:** We add the two factors found in step 1

<Tex block>
\begin{align*}
1+15&=16 \\\
3+5&=8 \\\
(-1)+(-15)&=-16 \\\
(-3)+(-5)&=-8
\end{align*}
</Tex>

Hence, <Tex block>\boxed{m=3\\; and\\; n=5}</Tex>

therefore, <Tex block>x^2+8x+15=(x+3)(x+5)</Tex>

<Example type="example" num="2"/>

Let's factoring the quadratic trinomial: <Tex>x^2-9x+20</Tex>

<Example type="solution"/>

<Tex>x^2-9x+20=(x+m)(x+n)</Tex>

**Step1:** We decompose the number 20:

<Tex block>
\begin{align*}
20&=1\times20 \\\
20&=4\times5 \\\
20&=(-1)\times(-20) \\\
20&=(-4)\times(-5)
\end{align*}
</Tex>

**Step2:** We add the two factors found in step 1

<Tex block>
\begin{align*}
1+20&=21 \\\
4+5&=9 \\\
(-1)+(-20)&=-21 \\\
(-4)+(-5)&=-9
\end{align*}
</Tex>

Hence, <Tex block>\boxed{m=-4\\; and\\; n=-5}</Tex>

therefore, <Tex block>x^2+8x+15=(x-4)(x-5)</Tex>

<Example type="example" num="3"/>

Let's factoring the quadratic trinomial: <Tex>x^2+4x-21</Tex>

<Example type="solution"/>

<Tex>x^2+4x-21=(x+m)(x+n)</Tex>

**Step1:** We decompose the number -21:

<Tex block>
\begin{align*}
-21&=(-1)\times21 \\\
-21&=(-3)\times7 \\\
-21&=1\times(-21) \\\
-21&=3\times(-7)
\end{align*}
</Tex>

**Step2:** We add the two factors found in step 1

<Tex block>
\begin{align*}
(-1)+21&=20 \\\
(-3)+7&=4 \\\
1+(-21)&=-20 \\\
3+(-7)&=-4
\end{align*}
</Tex>

Hence, <Tex block>\boxed{m=-3\\; and\\; n=7}</Tex>

therefore, <Tex block>x^2+4x-21=(x-3)(x+7)</Tex>

<Example type="example" num="4"/>

Let's factoring the quadratic trinomial: <Tex>x^4+6x^2+8</Tex>

<Example type="solution"/>

Let <Tex>X=x^2</Tex>

<Tex>x^4+6x^2+8=X^2+6X+8=(X+m)(X+n)</Tex>

**Step1:** We decompose the number 8:

<Tex block>
\begin{align*}
8&=1\times8 \\\
8&=2\times4 \\\
8&=(-1)\times(-8) \\\
8&=(-2)\times(-4)
\end{align*}
</Tex>

**Step2:** We add the two factors found in step 1

<Tex block>
\begin{align*}
1+8&=9 \\\
2+4&=6 \\\
(-1)+(-8)&=-9 \\\
(-2)+(-4)&=-6
\end{align*}
</Tex>

Hence, <Tex block>\boxed{m=2\\; and\\; n=4}</Tex>

therefore, <Tex block>X^2+6X+8=(X+2)(X+6)=(x^2+2)(x^2+6)</Tex>

<Example type="example" num="5"/>

Let's factoring the quadratic trinomial: <Tex>-x^2+5x-6</Tex>

<Example type="solution"/>

In this example, we have <Tex>a=-1</Tex>. We need to factor the trinomial by <Tex>-1</Tex>. By doing so, the signs of the three terms are switched

<Tex>-x^2+5x-6=-(x^2-5x+6)</Tex>

<Tex>-(x^2-5x+6)=-(x+m)(x+n)</Tex>

**Step1:** We decompose the number 6:

<Tex block>
\begin{align*}
6&=1\times6 \\\
6&=2\times3 \\\
6&=(-1)\times(-6) \\\
6&=(-2)\times(-3)
\end{align*}
</Tex>

**Step2:** We add the two factors found in step 1

<Tex block>
\begin{align*}
1+6&=7 \\\
2+3&=5 \\\
(-1)+(-6)&=-7 \\\
(-2)+(-3)&=-5
\end{align*}
</Tex>

Hence, <Tex block>\boxed{m=-2\\; and\\; n=-3}</Tex>

therefore, <Tex block>-x^2+5x-6=-(x^2-5x+6)=-(x-2)(x-3)</Tex>

<Alert type='warning'>

Note that not all quadratic trinomials can be factored to <Tex>(x+m)(x+n)</Tex>. For example, trinomial <Tex>x^2+x+1</Tex> does not factor.

</Alert>

<Related>

<Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/exponent-rules-properties" as="/lessons/intermediate-algebra/exponent-rules-properties">Exponent Notation: Definition and Rules</Link>

</Related>

Step by step, you will learn how to factoring trinomials when a=1 or a=-1.

Factoring a Simple Quadratic Trinomials

# Factoring Quadratics With a Leading Coefficient: <Tex>ax^2+bx+c</Tex>

<Related>

<Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials: x^2+bx+c</Link>

</Related>

You are undoubtedly familiar with the expression <Tex>ax^2+bx+c</Tex> and have already known it. In a previous lesson, we saw that the expression <Tex>ax^2+bx+c</Tex> is called a **quadratic trinomial**, and we saw how to factor this expression using a very easy method.

In this chapter, we will also try to work out the expression <Tex>ax^2+bx+c</Tex>, but in the case of <Tex>a\neq \plusmn1</Tex>, which is also an easy case, requiring only a little concentration. All you have to do is follow the basic steps which we will see in detail with many illustrative examples. In order to solidify this process, I suggest you re-do the examples again.

The method used in this lesson to factor the <Tex>ax^2+bx+c</Tex> trinomial is called the **ac method**.

## Steps To Factor Quadratic Trinomial with a Leading Coefficient

These steps are very simplified, you can skip some of them or adopt your own method when you master the skill of factoring.

**Step 1:**

At this stage, we multiply the basic factor <Tex>a</Tex> by the last factor <Tex>c</Tex>, and we get a number we call <Tex>k</Tex>. Then,

<Tex block>k=a \times c</Tex>

Then we decompose the number <Tex>k</Tex> into the product of two numbers: <Tex>m</Tex> and <Tex>n</Tex> so that <Tex>b=m+n</Tex>

**Step 2:**

At this step, after we have defined the two numbers <Tex>m</Tex> and <Tex>n</Tex> in the first stage, we substitute <Tex>b</Tex> by <Tex>m+n</Tex> into the expression <Tex>ax^2+bx+c</Tex>, and we get:

<Tex block>
\begin{align*}
ax^2+bx+c&=ax^2+(m+n)x+c \\\
&=ax^2+mx+nx+c
\end{align*}
</Tex>

**Step 3:**

At this step, we will group the numbers <Tex>ax^2+mx</Tex> and <Tex>nx+c</Tex> in two different brackets, between them the addition (+) operation (in some cases we will have to put a minus operation instead of the addition operation)

<Tex block>
\begin{align*}
ax^2+bx+c&=(ax^2+mx)+(nx+c)
\end{align*}
</Tex>

**Step 4:**

We factor the two expressions in parentheses and we get a common factor. If you do not get a common factor, then you have probably missed one of the above steps and must recalculate.

**Step 5:**

And finally, after we defined the common factor in the previous step, we now factor out the trinomial using this common factor. (This common factor must be a binomial expression in parentheses).

## Examples of How to Factor a Trinomial

<Example type="example" num="1"/>

Factor the trinomial <Tex>6x^2+11x-10</Tex> as a product of two binomials.

<Example type="solution"/>

**Step 1:**

We calculate the product <Tex>a \times c</Tex>. we have <Tex>k=a \times c = 6 \times (-10)=-60</Tex>

After we get the number <Tex>k=-60</Tex>, we re-decompose it into the product of two numbers where their sum is 10.

<Tex block>-60=-1\times60 \quad and \quad -1+60=59</Tex> 

<Tex block>-60=-2\times30 \quad and \quad -2+30=28</Tex>

<Tex block>-60=-3\times20 \quad and \quad -3+20=17</Tex>

<Tex block>-60=-4\times15 \quad and \quad \boxed{-4+15=11} \checkmark</Tex>

<Tex block>-60=-5\times12 \quad and \quad -5+12=7</Tex>

<Tex block>-60=-6\times10 \quad and \quad -6+10=4</Tex>

**Step 2:**

We substitute <Tex>11</Tex> by <Tex>(-4+15)</Tex>:

<Tex block>
\begin{align*}
6x^2+11x-10&=6x^2+(-4+15)x-10 \\\
&=6x^2-4x+15x-10
\end{align*}
</Tex>

**Step 3:**

Grouping the two numbers <Tex>6x^2-4x</Tex> and <Tex>15x-10</Tex> in different parentheses.

<Tex block>
\begin{align*}
6x^2+11x-10&=(6x^2-4x)+(15x-10)
\end{align*}
</Tex>

**Step 4:**

The prenthese <Tex>(6x^2-4x)</Tex> has the greatest common factor of <Tex>2x</Tex>, and the prenthese <Tex>(15x-10)</Tex> has the greatest common factor of <Tex>5</Tex>

<Tex block>
\begin{align*}
6x^2+11x-10&=(6x^2-4x)+(15x-10) \\\
&=(2x \times 3x-2x \times 2)+(5 \times 3x-5 \times 2) \\\
&=2x \colorbox{aqua}{$(3x-2)$} +5 \colorbox{aqua}{$(3x-2)$}
\end{align*}
</Tex>

**Step 5:**

We found that the common factor is <Tex>\colorbox{aqua}{$(3x-2)$}</Tex>, perform one last factorization to get to the final answer:

<Tex block>6x^2+11x-10=(2x+5)(3x-2)</Tex>

---------------

<Example type="example" num="2"/>

Factor the trinomial <Tex>8x^2+14x+3</Tex> as a product of two binomials.

<Example type="solution"/>

**Step 1:**

We calculate the product <Tex>a \times c</Tex>. we have <Tex>k=a \times c = 8 \times 3=24</Tex>

After we get the number <Tex>k=24</Tex>, we re-decompose it into the product of two numbers where their sum is 14.

<Tex block>24=1\times24 \quad and \quad 1+24=25</Tex> 

<Tex block>24=2\times12 \quad and \quad \boxed{2+12=14} \checkmark</Tex>

<Tex block>24=3\times8 \quad and \quad 3+8=11</Tex>

<Tex block>24=4\times6 \quad and \quad 4+6=10</Tex>

<Tex block>24=6\times10 \quad and \quad 6+10=16</Tex>

**Step 2:**

We substitute <Tex>14</Tex> by <Tex>(2+12)</Tex>:

<Tex block>
\begin{align*}
8x^2+14x+3&=8x^2+(2+12)x+3 \\\
&=8x^2+2x+12x+3
\end{align*}
</Tex>

**Step 3:**

Grouping the two numbers <Tex>8x^2+2x</Tex> and <Tex>12x+3</Tex> in different parentheses.

<Tex block>
\begin{align*}
8x^2+14x+3&=(8x^2+2x)+(12x+3)
\end{align*}
</Tex>

**Step 4:**

The prenthese <Tex>(8x^2+2x)</Tex> has the greatest common factor of <Tex>2x</Tex>, and the prenthese <Tex>(12x+3)</Tex> has the greatest common factor of <Tex>3</Tex>

<Tex block>
\begin{align*}
8x^2+14x+3&=(8x^2+2x)+(12x+3) \\\
&=(2x \times 4x+2x \times 1)+(3 \times 4x+3 \times 1) \\\
&=2x \colorbox{aqua}{$(4x+1)$} +3 \colorbox{aqua}{$(4x+1)$}
\end{align*}
</Tex>

**Step 5:**

We found that the common factor is <Tex>\colorbox{aqua}{$(4x+1)$}</Tex>, perform one last factorization to get to the final answer:

<Tex block>8x^2+14x+3=(2x+3)(4x+1)</Tex>

---------------

<Example type="example" num="3"/>

Factor the trinomial <Tex>12x^2-13x+3</Tex> as a product of two binomials.

<Example type="solution"/>

**Step 1:**

We calculate the product <Tex>a \times c</Tex>. we have <Tex>k=a \times c = 12 \times 3=36</Tex>

After we get the number <Tex>k=36</Tex>, we re-decompose it into the product of two numbers where their sum is -13.

<Tex block>36=1\times36 \quad and \quad 1+36=37</Tex> 

<Tex block>36=2\times18 \quad and \quad 2+18=20</Tex>

<Tex block>36=3\times12 \quad and \quad 3+12=15</Tex>

<Tex block>36=4\times9 \quad and \quad 4+9=13</Tex>

<Tex block>36=6\times6 \quad and \quad 6+6=12</Tex>

<Tex block>36=(-1)\times(-36) \quad and \quad (-1)+(-36)=-37</Tex> 

<Tex block>36=(-2)\times(-18) \quad and \quad (-2)+(-18)=-20</Tex>

<Tex block>36=(-3)\times(-12) \quad and \quad (-3)+(-12)=-15</Tex>

<Tex block>36=(-4)\times(-9) \quad and \quad \boxed{(-4)+(-9)=-13} \checkmark</Tex>

<Tex block>36=(-6)\times(-6) \quad and \quad (-6)+(-6)=-12</Tex>

**Step 2:**

We substitute <Tex>-13</Tex> by <Tex>((-4)+(-9))</Tex>:

<Tex block>
\begin{align*}
12x^2-13x+3&=12x^2+((-4)+(-9))x+3 \\\
&=12x^2-4x-9x+3
\end{align*}
</Tex>

**Step 3:**

Grouping the two numbers <Tex>12x^2-4x</Tex> and <Tex>-9x+3</Tex> in different parentheses.

<Tex block>
\begin{align*}
12x^2-13x+3&=(12x^2-4x)+(-9x+3) \\\
&=(12x^2-4x)-(9x-3)
\end{align*}
</Tex>

**Step 4:**

The prenthese <Tex>(12x^2-4x)</Tex> has the greatest common factor of <Tex>4x</Tex>, and the prenthese <Tex>(9x-3)</Tex> has the greatest common factor of <Tex>3</Tex>

<Tex block>
\begin{align*}
12x^2-13x+3&=(12x^2-4x)-(9x-3) \\\
&=(4x \times 3x-4x \times 1)-(3 \times 3x-3 \times 1) \\\
&=4x \colorbox{aqua}{$(3x-1)$} -3 \colorbox{aqua}{$(3x-1)$}
\end{align*}
</Tex>

**Step 5:**

We found that the common factor is <Tex>\colorbox{aqua}{$(3x-1)$}</Tex>, perform one last factorization to get to the final answer:

<Tex block>12x^2-13x+3=(4x-3)(3x-1)</Tex>

---------------

<Example type="example" num="5"/>

Factor the trinomial <Tex>3x^2+11x-14</Tex> as a product of two binomials.

<Example type="solution"/>

**Step 1:**

We calculate the product <Tex>a \times c</Tex>. we have <Tex>k=a \times c = 3 \times (-14)=-42</Tex>

After we get the number <Tex>k=-42</Tex>, we re-decompose it into the product of two numbers where their sum is 11.

<Tex block>-42=-1\times42 \quad and \quad -1+42=41</Tex> 

<Tex block>-42=-2\times21 \quad and \quad -2+21=19</Tex>

<Tex block>-42=-3\times14 \quad and \quad \boxed{-3+14=11} \checkmark</Tex>

<Tex block>-42=-6\times7 \quad and \quad -6+7=1</Tex>

**Step 2:**

We substitute <Tex>11</Tex> by <Tex>(-3+14)</Tex>:

<Tex block>
\begin{align*}
3x^2+11x-14&=3x^2+(-3+14)x-14 \\\
&=3x^2-3x+14x-14
\end{align*}
</Tex>

**Step 3:**

Grouping the two numbers <Tex>3x^2-3x</Tex> and <Tex>14x-14</Tex> in different parentheses.

<Tex block>
\begin{align*}
3x^2+11x-14&=(3x^2-3x)+(14x-14)
\end{align*}
</Tex>

**Step 4:**

The prenthese <Tex>(3x^2-3x)</Tex> has the greatest common factor of <Tex>3x</Tex>, and the prenthese <Tex>(14x-14)</Tex> has the greatest common factor of <Tex>14</Tex>

<Tex block>
\begin{align*}
3x^2+11x-14&=(3x^2-3x)+(14x-14) \\\
&=(3x \times x-3x \times 1)+(14 \times x-14 \times 1) \\\
&=3x \colorbox{aqua}{$(x-1)$} +14 \colorbox{aqua}{$(x-1)$}
\end{align*}
</Tex>

**Step 5:**

We found that the common factor is <Tex>\colorbox{aqua}{$(x-1)$}</Tex>, perform one last factorization to get to the final answer:

<Tex block>6x^2+11x-10=(3x+14)(x-1)</Tex>

Step by step, you will learn how to factoring quadratics with a leading coefficient (general case).

Factoring Quadratics With a Leading Coefficient

# Factoring Difference of Two Perfect Squares

<Related>

<Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">DSimplifying algebraic expressions by Combining Like Terms</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/exponent-rules-properties" as="/lessons/intermediate-algebra/exponent-rules-properties">Exponent Notation: Definition and Rules</Link>

</Related>

Factoring algebraic expressions are very important during your course of study to learn the basics of algebra. There are several different types of factoring cases, including using the greatest common factor (GCF), or using the grouping method, and also using special patterns where the factoring is very easy.

Among these special patterns, we find the **difference of two perfect squares**. By perfect squares, we mean two numbers raised to the power of 2.

<Image src="/imgs/lessons/intermalgebra/difference-squares.jpeg" alt="Difference squares terms" width={423} height={200} />

When we have an algebraic expression written as the difference of two perfect squares, we've guaranteed the factoring using the easiest way possible. So how do we factor that difference?

## Factoring Difference of Two Perfect Squares Formula

They are very familiar formulas, which can be simplified by the following scheme:

<Image src="/imgs/lessons/intermalgebra/difference-squares-formula.jpeg" alt="Difference squares terms" width={800} height={100} />

Now, let's move on to using the expression with variables:

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> two real numbers. We have:

<Tex block>a^2-b^2=(a+b)(a-b)</Tex>

</Note>

As a note, when we say that <Tex>a</Tex> and <Tex>b</Tex> are real numbers, this means that they can be known, or we can express them or one of them with another unknown, or one or both of them can be composite numbers.

For example:

<Tex block>x^2-y^2</Tex>

<Tex block>(6x)^2-y^2</Tex>

<Tex block>1-(\sqrt{5}x)^2</Tex>

<Tex block>(x+2)^2-4</Tex>

## Examples of How to Factor Difference of Two Perfect Squares

Here is a set of illustrative examples of how to factor the difference of two perfect squares. And we will guide you through all the steps, you can skip them when you are skilled enough and get the result straight away.

<Example type="example" num="1"/>

Factor the binomial below using the difference of two squares method.

<Tex block>a^2-25</Tex>

<Example type="solution"/>

As for the first term, it is clear that it is written in the form of a power raised to the exponent 2, but the second term which is 25 is not, so here we must look for a number if we raise it to the power of 2, we get 25, no doubt that you knew it quite easily because it is a very simple case which is <Tex>25=5^2</Tex>. So we will substitute <Tex>25</Tex> by <Tex>5^2</Tex> into the expression <Tex>a^2-25</Tex>, we get:

<Tex block>a^2-25=a^2-5^2</Tex>

We will take the base powers <Tex>a^2</Tex> and <Tex>5^2</Tex>, i.e. we take <Tex>a</Tex> and <Tex>5</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

<Tex block>
\begin{align*}
a^2-25&=a^2-5^2 \\\
&=(a+5)(a-5)
\end{align*}
</Tex>

-------------------------------

<Example type="example" num="2"/>

Factor the binomial below using the difference of two squares method.

<Tex block>81-36y^2</Tex>

<Example type="solution"/>

Both terms are not raised to the power of 2, here we must always show an exponent to be able to factor.

The number 81 is a perfect square of 9, i.e. <Tex>81=9^2</Tex>.

But the number <Tex>36y^2</Tex> is a product of two numbers, using power rules, we rewrite it as a perfect square: <Tex>36y^2=6^2 \times y^2=(6y)^2</Tex>.

Therefore,

<Tex block>81-36y^2=9-(6y)^2</Tex>

Same as the example 1, we will take the base powers <Tex>9^2</Tex> and <Tex>(6y)^2</Tex>, i.e. we take <Tex>9</Tex> and <Tex>6y</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

<Tex block>
\begin{align*}
81-36y^2&=9^2-(6y)^2 \\\
&=(9+6y)(9-6y)
\end{align*}
</Tex>

-------------------------------

<Example type="example" num="3"/>

Factor the binomial below using the difference of two squares method.

<Tex block>16a^2-9b^2</Tex>

<Example type="solution"/>

Both terms are not raised to the power of 2, here we must always show an exponent to be able to factor.

The number <Tex>16a^2</Tex> is a product of two numbers, using power rules, we rewrite it as a perfect square: <Tex>16a^2=16\times a^2=4^2 \times a^2=(4a)^2</Tex>.

And the number <Tex>9b^2</Tex> is a product of two numbers, using power rules, we rewrite it as a perfect square: <Tex>9b^2=9 \times b^2=3^2 \times b^2=(3b)^2</Tex>.

Therefore,

<Tex block>16a^2-9b^2=(4a)^2-(3b)^2</Tex>

Same as the examples above, we will take the base powers <Tex>(4a)^2</Tex> and <Tex>(3b)^2</Tex>, i.e. we take <Tex>4a</Tex> and <Tex>3b</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

<Tex block>
\begin{align*}
16a^2-9b^2&=(4a)^2-(3b)^2 \\\
&=(4a+3b)(4a-3b)
\end{align*}
</Tex>

-------------------------------

<Example type="example" num="4"/>

Factor the binomial below using the difference of two squares method.

<Tex block>(x+5)^2-49</Tex>

<Example type="solution"/>

Both terms are not raised to the power of 2, here we must always show an exponent to be able to factor.

The number <Tex>49</Tex> is a perfect square of 7, i.e, <Tex>49=7^2</Tex>.

And the number <Tex>(x+5)^2</Tex> is raised to the power of 2.

Therefore,

<Tex block>(x+5)^2-49=(x+5)^2-7^2</Tex>

Same as the examples above, we will take the base powers <Tex>(x+5)^2</Tex> and <Tex>7^2</Tex>, i.e. we take <Tex>(x+5)</Tex> and <Tex>7</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

**Note:** You must simplify the expressions inside the parentheses by removing nested parentheses then applying combine like terms.

<Tex block>
\begin{align*}
(x+5)^2-49&=(x+5)^2-7^2 \\\
&=((x+5)+7)((x+5)-7) \\\
&=(x+5+7)(x+5-7) \\\
&=(x+12)(x-2)
\end{align*}
</Tex>

-------------------------------

<Example type="example" num="5"/>

Factor the binomial below using the difference of two squares method.

<Tex block>a^4-b^4</Tex>

<Example type="solution"/>

Both terms of the expression <Tex>a^4-b^4</Tex> are raised to the power of 4, wich mean that we can rewrite it in order to obtain the exponent 2. Note that all powers that the exponent is even we can rewrite it as power of 2.

Therefore,

<Tex block>a^4-b^4=(a^2)^2-(b^2)^2</Tex>

Same as the examples above, we will take the base powers <Tex>(a^2)^2</Tex> and <Tex>(b^2)^2</Tex>, i.e. we take <Tex>a^2</Tex> and <Tex>b^2</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

<Tex block>
\begin{align*}
(a^2)^2-(b^2)^2&=(a^2)^2-(b^2)^2 \\\
&=(a^2+b^2)(a^2-b^2) \\\
\end{align*}
</Tex>

You may think we are done the factoring, no, my friend, notice carefully the last pair of brackets, does it remind you of the expression <Tex>(a^2-b^2)</Tex>? Yes, it is a difference of two perfect squares, which means that it is factorable. Therefore,

<Tex block>a^2-b^2=(a+b)(a-b)</Tex>

Hence,

<Tex block>
\begin{align*}
a^4-b^4&=(a^2+b^2)(a^2-b^2) \\\
&=(a^2+b^2)(a+b)(a-b)
\end{align*}
</Tex>

-------------------------------

<Example type="example" num="6"/>

Factor the binomial below using the difference of two squares method.

<Tex block>x^2-3</Tex>

<Example type="solution"/>

As for the first term, it is clear that it is written in the form of a power raised to the exponent 2, but the second term which is 3 is not, so here we must look for a number if we raise it to the power of 2, we get 3. What is the number multiplied by itself to get 3? I hope you know it: <Tex>3=\sqrt3^2</Tex>. So we will substitute <Tex>3</Tex> by <Tex>\sqrt3^2</Tex> into the expression <Tex>x^2-3</Tex>, we get:

<Tex block>x^2-3=x^2-\sqrt3^2</Tex>

This formula seems familiar to you, yes it is exactly what we wrote in the previous rule. So we will take the base powers <Tex>x^2</Tex> and <Tex>\sqrt3^2</Tex>, i.e. we take <Tex>x</Tex> and <Tex>\sqrt3</Tex>, and then put them inside two pairs of parentheses, where there will be two opposite signs between the two numbers as follows:

<Tex block>
\begin{align*}
x^2-3&=x^2-\sqrt3^2 \\\
&=(x+\sqrt3)(x-\sqrt3)
\end{align*}
</Tex>

Step by step, you will learn how to factoring difference of two perfect squares (a^2-b^2=(a+b)(a-b)) illustrated by many examles with solution.

Factoring Difference of Two Perfect Squares

# FOIL Method For Multiplying Two Binomials

FOIL Method is a technique that enables us to multiply two binomials in a very simple and easy way. But before explaining this method, it is necessary to recall the definition of binomial. A binomial is a special case of a polynomial, which is an algebraic expression consisting of only two terms separated by one of the + or - signs.

<Image src="/imgs/lessons/intermalgebra/foil_method.webp" alt="FOIL method for distribute two binomials" width='551px' height='268px'/>

## FOIL method mathematically

Foil method is a technique that makes it easier for a beginner student to remember the steps of the expanding process for an algebraic expression consisting of the product of two binomial terms written as follows:

<Tex block>(a+b)(c+d)</Tex>

The F-O-I- L acronym stands for First, Outer, Inner, and Last.

Let’s explain each of these terms:

* F: the first, that is, we multiply the first two terms from each parenthesis:

<Tex block>(\bold{a}+b)(\bold{c}+d)=\bold{ac}+\dots</Tex>

* O: the outers, ths is, we multiply the outermost terms of the two binomials:

<Tex block>(\bold{a}+b)(c+\bold{d})=ac+\bold{ad} \dots</Tex>

* I: the inners, that is, we multiply the innermost terms of the two binomials:

<Tex block>(a+\bold{b})(\bold{c}+d)=ac+ad+\bold{bc} \dots</Tex>

* L: the last, that is, we multiply the last two terms from each parenthesis:

<Tex block>(a+\bold{b})(c+\bold{d})=ac+ad+bc+\bold{bd}</Tex>

When we get the four sums, we simplify like terms by Combining Like Terms.

<Note>

Let <Tex>a</Tex>, <Tex>b</Tex>, <Tex>c</Tex> and <Tex>d</Tex> a real numbers.

<Tex block>(a+b)(c+d)=ac+ad+bc+bd</Tex>

</Note>

<Alerts type='warning'>

Pay attention to the sign of the numbers <Tex>a</Tex>, <Tex>b</Tex>, <Tex>c</Tex> and <Tex>d</Tex> in the above formula.

</Alert>

## Examples of How to Multiply Binomials using the FOIL Method

<Example type="example" num="1"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x+2)(x+3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(x+3)&=x \times x+\dots \\\
&=x^2+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(x+3)&=x \times x+x \times3\dots \\\
&=x^2+3x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(x+3)&=x \times x+x \times3+2\times x\dots \\\
&=x^2+3x+2x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(x+3)&=x \times x+x \times3+2\times x+2\times3 \\\
&=x^2+3x+2x+6
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>3x</Tex> and <Tex>2x</Tex> are like terms, hence:

<Tex block>(x+2)(x+3)=x^2+5x+6</Tex>

-------------

<Example type="example" num="2"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x-2)(x+3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x-2)(x+3)&=x \times x+\dots \\\
&=x^2+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x-2)(x+3)&=x \times x+x \times3\dots \\\
&=x^2+3x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x-2)(x+3)&=x \times x+x \times3-2\times x\dots \\\
&=x^2+3x-2x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x-2)(x+3)&=x \times x+x \times3-2\times x-2\times3 \\\
&=x^2+3x-2x-6
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>3x</Tex> and <Tex>-2x</Tex> are like terms, hence:

<Tex block>(x-2)(x+3)=x^2+x+6</Tex>

--------------

<Example type="example" num="3"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x+2)(x-3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(x-3)&=x \times x+\dots \\\
&=x^2+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(x-3)&=x \times x-x \times3\dots \\\
&=x^2-3x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(x-3)&=x \times x-x \times3+2\times x\dots \\\
&=x^2-3x+2x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(x-3)&=x \times x+x \times3+2\times x-2\times3 \\\
&=x^2-3x+2x-6
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>-3x</Tex> and <Tex>2x</Tex> are like terms, hence:

<Tex block>(x+2)(x-3)=x^2-x-6</Tex>

-------------

<Example type="example" num="4"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x-2)(x-3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x-2)(x-3)&=x \times x+\dots \\\
&=x^2+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x-2)(x-3)&=x \times x-x \times3\dots \\\
&=x^2-3x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x-2)(x-3)&=x \times x-x \times3-2\times x\dots \\\
&=x^2-3x-2x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x-2)(x-3)&=x \times x-x \times3-2\times x+2\times3 \\\
&=x^2-3x-2x+6
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>-3x</Tex> and <Tex>-2x</Tex> are like terms, hence:

<Tex block>(x+2)(x-3)=x^2-5x+6</Tex>

-------------

<Example type="example" num="5"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(2x+1)(5x+3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(2x+1)(5x+3)&=2x \times 5x+\dots \\\
&=10x^2+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(2x+1)(5x+3)&=2x \times 5x+2x \times3\dots \\\
&=10x^2+6x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(2x+1)(5x+3)&=2x \times 5x+2x \times3+1\times 5x\dots \\\
&=10x^2+6x+5x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(2x+1)(5x+3)&=2x \times 5x+2x \times3+1\times 5x+1\times3 \\\
&=10x^2+6x+5x+3
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>6x</Tex> and <Tex>5x</Tex> are like terms, hence:

<Tex block>(2x+1)(5x+3)=10x^2+11x+3</Tex>

-------------

<Example type="example" num="6"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x^2+5)(x-7)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x^2+5)(x-7)&=x^2 \times x+\dots \\\
&=x^3+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x^2+5)(x-7)&=x^2 \times x-x^2 \times7\dots \\\
&=x^3-7x^2+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x^2+5)(x-7)&=x^2 \times x-x^2 \times7+5\times x\dots \\\
&=x^3-7x^2+5x+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x^2+5)(x-7)&=x^2 \times x-x^2 \times7+5\times x-5\times7 \\\
&=x^3-7x^2+5x-35
\end{align*}
</Tex>

We don't have like terms in this expression, hence:

<Tex block>(x^2+5)(x-7)=x^3-7x^2+5x-35</Tex>

-------------

<Example type="example" num="7"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(1-5x)(x+8)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(1-5x)(x+8)&=1 \times x+\dots \\\
&=x+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(1-5x)(x+8)&=1 \times x+1 \times8\dots \\\
&=x+8+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(1-5x)(x+8)&=1 \times x+1 \times8-5x\times x\dots \\\
&=x+8-5x^2+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(1-5x)(x+8)&=1 \times x+1 \times8-5x\times x-5x\times8 \\\
&=x+8-5x^2-40x
\end{align*}
</Tex>

Having applied the FOIL method, now we simplify the expression obtained using <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms.</Link>

We have: <Tex>x</Tex> and <Tex>-40x</Tex> are like terms, hence:

<Tex block>(1-5x)(x+8)=-5x^2-39x+8</Tex>

------------------

<Example type="example" num="1"/>

Distribute and simplify the expression below using FOIL method:

<Tex block>(x+2)(y+3)</Tex>

<Example type="solution"/>

* First: multiply the first two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(y+3)&=x \times y+\dots \\\
&=xy+\dots
\end{align*}
</Tex>

* Outer: we multiply the outermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(y+3)&=x \times y+x \times3\dots \\\
&=xy+3x+\dots
\end{align*}
</Tex>

* Inner: we multiply the innermost terms of the two binomials:

<Tex block>
\begin{align*}
(x+2)(y+3)&=x \times y+x \times3+2\times y\dots \\\
&=xy+3x+2y+\dots
\end{align*}
</Tex>

* Last: multiply the last two terms from each binomial:

<Tex block>
\begin{align*}
(x+2)(y+3)&=x \times y+x \times3+2\times y+2\times3 \\\
&=xy+3x+2y+6
\end{align*}
</Tex>

We don't have like terms in this expression, hence:

<Tex block>(x+2)(y+3)=xy+3x+2y+6</Tex>

Step by step, you will learn how to expanding and multiplying two binomials with double brackets using foil method math.

Foil Method For Multiplying Binomials

# Find Roots Of Quadratic Equation Using Quadratic Formula

A Quadratic Equation is a trinomial (Polynomial with the third degree). To solve such equations, there are a few methods. For example, Completing the square, factoring. But the most popular method used to solve such equations is the method of calculating the discriminant of the equation - the so-called quadratic formula. This formula, as we will see later, gives us an idea about whether this equation has real solutions or does not. And the number of solutions if it has solutions.

## The standard form of the quadratic equation

In a previous lesson, we saw that <Link to="../basic-algebra/solving-simple-linear-equation-in-one-variable" as="/lessons/basic-algebra/solving-simple-linear-equation-in-one-variable">linear equations</Link> are written in the form of x, where x and x are two known numbers and x is an unknown number. Solving this equation requires finding the value of the unknown x that makes both sides of the equation equal.

For Quadratic Equations, the matter is slightly different from linear equations, since the unknown, in this case, is raised to the power of 2, and therefore the method of solving will differ from linear equations.

Here's the definition of a quadratic equations.

<Note>
Let <Tex>a</Tex>, <Tex>b</Tex> and <Tex>c</Tex> represent real numbers with <Tex>a\neq0</Tex>. The standard form of the quadratic equations is:

<Tex block>\boxed{ax^2+bx+c=0}</Tex>

</Note>

<Image src="/imgs/lessons/intermalgebra/equation_quadratic_example.png" alt="examples of quadratic equations with location of coefficients" width='706px' height='579px'/>

## The Discriminant of a Quadratic Equation

At the beginning of this lesson, we said that the discriminant of the equation determines a very important thing, which is the number of solutions to the equation. So how do we calculate this important mathematical expression? The formula is very simple, you just have to know well the positions of the coefficients <Tex>a</Tex>, <Tex>b</Tex> and <Tex>c</Tex> in the equation.

Here is the rule we use for calculate the discriminant.

<Note>

We consider the following quadratic equation.

<Tex block>ax^2+bx+c=0\qquad a\neq0</Tex>

The discriminant of this equation is:

<Tex block>\Delta=b^2-4ac</Tex>

</Note>

## The Roots of a Quadratic Equation Using the Quadratic Formula

The discriminant of the equation can therefore be a positive, negative or equal to 0. The sign of this discriminant is what will determine the number of solutions to the equation which are three different cases detailed in the following property.

<Note>

- If <Tex>\Delta >0</Tex>. The equation has two different real roots:

<Tex block>x=\dfrac{-b-\sqrt{\Delta}}{2a}\quad or \quad x=\dfrac{-b+\sqrt{\Delta}}{2a}</Tex>

In short, we write:

<Tex block>x=\dfrac{-b \plusmn \sqrt{\Delta}}{2a}</Tex>

- If <Tex>\Delta =0</Tex>. The equation has one real root:

<Tex block>x=\dfrac{-b}{2a}</Tex>

- If <Tex>\Delta <0</Tex>. The equation has no real roots:

</Note>

_______

## Examples of Using the Quadratic Formula to Solve Quadratic Equations

In these examples we will try to take some examples from the three cases to solve a quadratic equation.

**Steps to solve the quadratic equation:**

- Locate the coefficients <Tex>a</Tex>, <Tex>b</Tex> and <Tex>c</Tex> (Check the table above).

- Calculate <Tex>\Delta</Tex>.

- Check the sign of <Tex>\Delta</Tex>.

- Derive the solutions to the equation from the above property.

<Example type="example" num="1"/>

Solve the equation: <Tex>x^2-7x-8=0</Tex>

<Example type="solution"/>

- The coefficients:

<Tex block>a=1 \quad ; \quad b=-7 \quad ; \quad c=-8</Tex>

- Calculate <Tex>\Delta</Tex>:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= (-7)^2-4\times1\times(-8)\\\
&= 49+81\\\
&= 91
\end{align*}
</Tex>.

- The sign of <Tex>\Delta</Tex>:

<Tex block>\Delta=81>0</Tex>

So, there two real roots:

<Tex block>
\begin{align*}
x&=\dfrac{-b\plusmn \sqrt{\Delta}}{2a}\\\
&= \dfrac{-(-7)\plusmn \sqrt{81}}{2\times1}\\\
&= \dfrac{7\plusmn9}{2}\\\
x&=\dfrac{7-9}{2} \quad or \quad x=\dfrac{7+9}{2}\\\
x&=\dfrac{-2}{2} \quad or \quad x=\dfrac{16}{2}\\\
x&=-1 \quad or \quad x=8
\end{align*}
</Tex>

Hence, the two roots are: <Tex>x=-1 \quad or \quad x=8</Tex>

<Example type="example" num="2"/>

Solve for x the equation: <Tex>3x^2+7x-4=2</Tex>

<Example type="solution"/>

The equation is not in the standard form, because the left side is different from 0. So, we must make this equation in standard form by subtracting 2 from both sides.

<Tex block>3x^2+7x-4=2</Tex>

is equivalent to: <Tex>3x^2+7x-4-2=2-2</Tex>

is equivalent to: <Tex>\boxed{3x^2+7x-6=0}</Tex>

- The coefficients:

<Tex block>a=3 \quad ; \quad b=7 \quad ; \quad c=-6</Tex>

- Calculate <Tex>\Delta</Tex>:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 7^2-4\times3\times(-6)\\\
&= 49+72\\\
&= 121
\end{align*}
</Tex>.

- The sign of <Tex>\Delta</Tex>:

<Tex block>\Delta=121>0</Tex>

So, there two real roots:

<Tex block>
\begin{align*}
x&=\dfrac{-b\plusmn \sqrt{\Delta}}{2a}\\\
&= \dfrac{-7\plusmn \sqrt{121}}{2\times3}\\\
&= \dfrac{7\plusmn11}{6}\\\
x&=\dfrac{7-11}{6} \quad or \quad x=\dfrac{7+11}{6}\\\
x&=\dfrac{-4}{6} \quad or \quad x=\dfrac{18}{6}\\\
x&=-\dfrac{2}{3} \quad or \quad x=3
\end{align*}
</Tex>

Hence, the two roots are: <Tex>x=-\dfrac{2}{3} \quad or \quad x=3</Tex>

<Example type="example" num="3"/>

Solve for x the equation: <Tex>x^2+4x+4=0</Tex>

<Example type="solution"/>

- The coefficients:

<Tex block>a=1 \quad ; \quad b=4 \quad ; \quad c=4</Tex>

- Calculate <Tex>\Delta</Tex>:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 4^2-4\times1\times4\\\
&= 16-16\\\
&= 0
\end{align*}
</Tex>.

- The sign of <Tex>\Delta</Tex>:

<Tex block>\Delta=0</Tex>

So, there one real root:

<Tex block>
\begin{align*}
x&=\dfrac{-b}{2a}\\\
&= \dfrac{-4}{2\times1}\\\
&= \dfrac{-4}{2}\\\
x&=-2
\end{align*}
</Tex>

Hence, the one solution is: <Tex>x=-2</Tex>

<Example type="example" num="4"/>

Solve for x the equation: <Tex>x^2+x+1=0</Tex>

<Example type="solution"/>

- The coefficients:

<Tex block>a=1 \quad ; \quad b=1 \quad ; \quad c=1</Tex>

- Calculate <Tex>\Delta</Tex>:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 1^2-4\times1\times1\\\
&= 1-4\\\
&= -3
\end{align*}
</Tex>.

- The sign of <Tex>\Delta</Tex>:

<Tex block>\Delta=-3<0</Tex>

So, there are no real solutions for this quadratic equation.

-----------------

<Related>

<Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring Difference of Two Perfect Squares</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials</Link> 

<Link to="../intermediate-algebra/factoring-trinomials-using-quadratic-equation-roots" as="/lessons/intermediate-algebra/factoring-trinomials-using-quadratic-equation-roots">Factoring Trinomials Using Quadratic Equation Roots</Link>

</Related>

Learn how to find roots (zeros) of quadratic equation (trinomial quadratic) using quadratic formula and the discriminant with many illustrated examples.

Find Roots Of Quadratic Equation Using Quadratic Formula

# Factoring Trinomials Using Quadratic Equation Roots

<Image src="/imgs/lessons/intermalgebra/factoring_trinomials.jpg" alt="factoring trinomials using quadratic formula" width='640px' height='360px' />

For factorization of trinomials, there are several different ways including:

- Factoring by grouping *(will be discussed later)*;

- Factoring by completing the square *(will be discussed later)*;

- Factoring a <Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">simple trinomial</Link>;

- Factoring with the <Link to="../intermediate-algebra/factoring-quadratics-with-leading-coefficient" as="/lessons/intermediate-algebra/factoring-quadratics-with-leading-coefficient">leading coefficient</Link>;

But in this lesson, we will show another method, which may be the best in the event that it is not possible to use the other methods, and it may also be the easiest.

This method depends on transforming the trinomial into a quadratic equation, then we solve this equation as explained in the lesson <Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">"Solving of a Quadratic Equations"</Link>. When we find the solutions to the equation, we can then infer factoring the trinomial.

The full method will be explained, just follow the article.

## Transforming the Trinomial into a Quadratic Equation and Solve it

To factor a trinomial, first we must transform it to a standard form of a quadratic equation.

<Note>

Given a trinomial: <Tex>ax^2+bx+c</Tex> with <Tex>a\neq0</Tex>.

The equivalent quadratic equation is: <Tex>ax^2+bx+c=0</Tex>

</Note>

In the standard form, there is always 0 on the right side, and the leading coefficient <Tex></Tex> must be different from 0.

To solve this equation, we must calculate the discriminant of the equation.

<Tex block>\Delta =b^2-4ac</Tex>

## Factoring Trinomials Formula

The quantity <Tex>\Delta</Tex> can be positive, negative, or equal to 0.

- If <Tex>\Delta >0 </Tex>, so, there are two real roots;

- If <Tex>\Delta =0 </Tex>, so, there is one real root;

- If <Tex>\Delta <0 </Tex>, so, there are no real roots;

The note bellow show you how to factor the trinomial according to the roots.

<Note>

Let <Tex>ax^2+bx+c=0</Tex> be a quadratic equation.

- If <Tex>\Delta >0</Tex>. The equation has two different real roots:

<Tex block>x_1=\dfrac{-b-\sqrt{\Delta}}{2a}\quad or \quad x_2=\dfrac{-b+\sqrt{\Delta}}{2a}</Tex>

Rewriting the trinomial as a factoring form:

<Tex block>\boxed{ax^2+bx+c=a(x-x_1)(x-x_2)}</Tex>

- If <Tex>\Delta =0</Tex>. The equation has one real root:

<Tex block>x_1=\dfrac{-b}{2a}</Tex>

Rewriting the trinomial as a factoring form:

<Tex block>\boxed{ax^2+bx+c=a(x-x_1)^2}</Tex>

- If <Tex>\Delta <0</Tex>. The equation has one real root. Then the trinomial cannot be factored.

</Note>

Note that <Tex>a</Tex> is the leading coefficient of the trinomial.

## Examples of factoring Trinomials Using Quadratic Formula

In these examples we will try to take some examples from the three cases to factor a trinomials using quadratic formula.

<Example type="example" num="1"/>

Factor the trinomial: <Tex>x^2-7x-8</Tex>

<Example type="solution"/>

- The equivalent quadratic equation is:

<Tex block>x^7-7x-8=0</Tex>

- Calculate the discriminant:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= (-7)^2-4\times1\times(-8)\\\
&= 49+32\\\
&= 81 >0
\end{align*}
</Tex>

- The discriminant is positive, so, there are two real roots:

<Tex block>
\begin{align*}
x_1&=\dfrac{-b-\sqrt{\Delta}}{2a}\quad and \quad x_2=\dfrac{-b+\sqrt{\Delta}}{2a}\\\
x_1&=\dfrac{-(-7)-\sqrt{81}}{2\times1}\quad and \quad x_2=\dfrac{-(-7)+\sqrt{81}}{2\times1}\\\
x_1&=\dfrac{7-9}{2}\quad and \quad x_2=\dfrac{7+9}{2}\\\
x_1&=\dfrac{-2}{2}\quad and \quad x_2=\dfrac{16}{2}\\\
x_1&=-1 \quad and \quad x_2=8
\end{align*}
</Tex>

- The trinomial can be factored as follow:

<Tex block>
\begin{align*}
x^2-7x-8 &=a(x-x_1)(x-x_2)\\\
&=1(x-(-1))(x-8)\\\
&=(x+1)(x-8)
\end{align*}
</Tex>

<Example type="example" num="2"/>

Factor the trinomial: <Tex>2x^2+9x-5</Tex>

<Example type="solution"/>

- The equivalent quadratic equation is:

<Tex block>2x^2+9x-5=0</Tex>

- Calculate the discriminant:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 9^2-4\times2\times(-5)\\\
&= 81+40\\\
&= 121 >0
\end{align*}
</Tex>

- The discriminant is positive, so, there are two real roots:

<Tex block>
\begin{align*}
x_1&=\dfrac{-b-\sqrt{\Delta}}{2a}\quad and \quad x_2=\dfrac{-b+\sqrt{\Delta}}{2a}\\\
x_1&=\dfrac{-9-\sqrt{121}}{2\times2}\quad and \quad x_2=\dfrac{-9+\sqrt{121}}{2\times2}\\\
x_1&=\dfrac{-9-11}{4}\quad and \quad x_2=\dfrac{-9+11}{4}\\\
x_1&=\dfrac{-20}{4}\quad and \quad x_2=\dfrac{2}{4}\\\
x_1&=-5 \quad and \quad x_2=\dfrac{1}{2}
\end{align*}
</Tex>

- The trinomial can be factored as follow:

<Tex block>
\begin{align*}
2x^2+9x-5 &=a(x-x_1)(x-x_2)\\\
&=2(x-(-5))(x-\dfrac{1}{2})\\\
&=2(x+5)(x-\dfrac{1}{2})
\end{align*}
</Tex>

<Example type="example" num="3"/>

Factor the trinomial: <Tex>-x^2+6x-9</Tex>

<Example type="solution"/>

- The equivalent quadratic equation is:

<Tex block>-x^2+6x-9=0</Tex>

- Calculate the discriminant:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 6^2-4\times(-1)\times(-9)\\\
&= 36-36\\\
&= 0
\end{align*}
</Tex>

- The discriminant is equal to zero, so, there is one real root:

<Tex block>
\begin{align*}
x_1&=\dfrac{-b}{2a}\\\
x_1&=\dfrac{-6}{2\times(-1)}\\\
x_1&=\dfrac{-6}{-2}\\\
x_1&=3
\end{align*}
</Tex>

- The trinomial can be factored as follow:

<Tex block>
\begin{align*}
-x^2+6x-9 &=a(x-x_1)^2\\\
&=-1\times(x-3)^2\\\
&=-(x-3)^2
\end{align*}
</Tex>

<Example type="example" num="4"/>

Factor the trinomial: <Tex>4x^2+8x+4</Tex>

<Example type="solution"/>

- The equivalent quadratic equation is:

<Tex block>4x^2+8x+4=0</Tex>

- Calculate the discriminant:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 8^2-4\times4\times4\\\
&= 64-64\\\
&= 0
\end{align*}
</Tex>

- The discriminant is equal to zero, so, there is one real root:

<Tex block>
\begin{align*}
x_1&=\dfrac{-b}{2a}\\\
x_1&=\dfrac{-8}{2\times4}\\\
x_1&=\dfrac{-8}{8}\\\
x_1&=-1
\end{align*}
</Tex>

- The trinomial can be factored as follow:

<Tex block>
\begin{align*}
4x^2+8x+4 &=a(x-x_1)^2\\\
&=4\times(x-(-1))^2\\\
&=4(x+1)^2
\end{align*}
</Tex>

<Example type="example" num="5"/>

Factor the trinomial: <Tex>3x^2+2x+1</Tex>

<Example type="solution"/>

- The equivalent quadratic equation is:

<Tex block>3x^2+2x+1=0</Tex>

- Calculate the discriminant:

<Tex block>
\begin{align*}
\Delta &= b^2-4ac\\\
&= 2^2-4\times3\times1\\\
&= 4-12\\\
&= -8\<0
\end{align*}
</Tex>

- The discriminant is negative, so, there are no real roots, and the trinomial cannot be factored.

-----------------

<Related>

<Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring Difference of Two Perfect Squares</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials</Link> 

<Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">Find Roots Of Quadratic Equation Using Quadratic Formula</Link>

</Related>

Learn how to factor trinomials (quadratic trinomials) by solving a quadratic equation and derive the factoring from the roots of the equation.

Factoring Trinomials Using Quadratic Equation Roots

# Factoring Sum and Difference of Two Perfect Cubes

We saw how to <Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">factor difference of two perfect squares</Link> using the following formula:

<Tex block>\boxed{\Large a^2-b^2=(a-b)(a+b)}</Tex>

In this lesson, we will try to identify how to factor the difference and sum of two perfect cubes using a very specific pattern.

But first, let's get to know what a cube is.

A cube is a number that can be converted to a power raised to the exponent of 3.

**List of first 11 perfect cube numerical values:**

<Text block>
\begin{align*}
0^3&=0\\\
1^3&=1\\\
2^3&=8\\\
3^3&=27\\\
4^3&=64\\\
5^3&=125\\\
6^3&=216\\\
7^3&=343\\\
8^3&=512\\\
9^3&=729\\\
10^3&=1000
\end{align*}
</Tex>

**Algebraic expressions for perfect cubes:**

<Text block>x^3\\; ;\\; x^6\\;;\\; x^9\\;;\\; x^{12}\\;;\\; x^{15}\\;;\\; x^{18}\\;;\\; x^{21}</Tex>

Note that each power with an exponent of multiples of 3 is also a perfect cube number.

<Text block>x^3\\; ;\\; (x^2)^3\\;;\\; (x^3)^3\\;;\\; (x^4)^3\\;;\\; (x^5)^3\\;;\\; (x^6)^3\\;;\\; (x^7)^3</Tex>

The formula in which the sum and difference of perfect squares are factored:

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> represent two real numbers.

Factoring the sum of two perfect cubes:

<Tex block>\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}</Tex>

Factoring the difference of two perfect cubes:

<Tex block>\boxed{a^3-b^3=(a-b)(a^2+ab+b^2)}</Tex>

</Note>

There is a simple way to help you remember these patterns. Note carefully in the formula factored, the first binomial retains the same sign as between the first and second terms of the left-hand side. While a trinomial, we always put between the first term and the second term the opposite of the sign as between the first and second terms of the left-hand side. Finally, the third term is always preceded by a + sign.

<Image src="/imgs/lessons/intermalgebra/factoring_cubes.jpeg" alt="factoring sum and difference of perfect cubes" width={640} height={500} />

Now let's take a variety of illustrations to factor out the addition and difference of two perfect cubes.

__________

## Examples of Factoring Sum and Difference of Two Perfect Cubes.

<Example type="example" num="1"/>

Factor the difference: <Tex>x^3-27</Tex>

<Example type="solution"/>

This algebraic expression is written as a difference. We notice that the first number is raised to the power of 3. The number 27 is not. So, to factor this algebraic expression, it is necessary to convert the number 27 to a power of 3. What is this power then?

No doubt you know the answer, the number <Tex>27=3\times9</Tex>, if we decompose the number 9 we get <Tex>27=3\times3\times3</Tex>, so <Tex>27=3^3</Tex>

Now we get the difference of two perfect cubes. Thus, the expression can be factored using the second form in the rule above.

<Tex block>
\begin{align*}
x^3-27&=x^3-3^3\\\
&=(x-3)(x^2+x\times3+3^2)\\\
&=(x-3)(x^2+3x+9)
\end{align*}
</Tex>

<Example type="example" num="2"/>

Factor the sum: <Tex>125+y^3</Tex>

<Example type="solution"/>

In the same way, this algebraic expression is written as a sum. We notice that the second number is raised to the power of 3. The number 125 is not. So, to factor this algebraic expression, it is necessary to convert the number 125 to a power of 3. What is this power then?

The number <Tex>125=5\times25</Tex>, if we decompose the number 25 we get <Tex>125=5\times5\times5</Tex>, so <Tex>125=5^3</Tex>

Now we get the sum of two perfect cubes. Thus, the expression can be factored using the first form in the rule above.

<Tex block>
\begin{align*}
125+y^3&=5^3+y^3\\\
&=(5+y)(5^2-5\times y+y^2)\\\
&=(5+y)(25-5y+y^2)
\end{align*}
</Tex>

<Example type="example" num="3"/>

Factor the sum: <Tex>8x^3+y^3</Tex>

<Example type="solution"/>

Rewrite this expression to obtain the sum of two perfect powers. Note that the number 8 is a perfect cube.

<Tex>8x^3=2^3\times x^3=(2x)^3</Tex>

Now we get the sum of two perfect cubes. Thus, the expression can be factored using the first form in the rule above.

<Tex block>
\begin{align*}
8x^3+y^3&=(2x)^3+y^3\\\
&=(2x+y)((2x)^2-2x\times y+y^2)\\\
&=(2x+y)(4x^2-2xy+y^2)
\end{align*}
</Tex>

<Example type="example" num="4"/>

Factor the difference: <Tex>64a^3-216b^3</Tex>

<Example type="solution"/>

Rewrite this expression to obtain the difference of two perfect powers. Note that the two numbers 64 and 216 are a perfect cubes.

<Tex>64a^3=4^3\times a^3=(4a)^3</Tex>

and <Tex>216b^3=6^3\times b^3=(6b)^3</Tex>

Now we get the difference of two perfect cubes. Thus, the expression can be factored using the second form in the rule above.

<Tex block>
\begin{align*}
64a^3-216b^3&=(4a)^3-(6b)^3\\\
&=(4a-6b)((4a)^2+4a\times 6b+(6b)^2)\\\
&=(4a-6b)(16x^2+24ab+36b^2)
\end{align*}
</Tex>

But the the expression obtained is not fully factored. Using factoring out <Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">the Greatest Common Factor</Link>:

- In the binomial, <Tex>gcf(4,6)=2</Tex>, Thus,

<Tex block>4a-6b=2(2a-3b)</Tex>

- In the trinomial, <Tex>gcf(16,24,36)=4</Tex>, Thus,

<Tex block>16x^2+24ab+36b^2=4(4x^2+6ab+9b^2)</Tex>

Therefore, 

<Tex block>
\begin{align*}
64a^3-216b^3&=(4a-6b)(16x^2+24ab+36b^2)\\\
&=2(2a-3b)\times4(4a^2+6ab+9b^2)\\\
&=8(2a-3b)(4a^2+6ab+9b^2)
\end{align*}
</Tex>

<Example type="example" num="5"/>

Factor the sum: <Tex>x^6+1</Tex>

<Example type="solution"/>

Rewrite this expression to obtain the sum of two perfect powers. Note that the <Tex>x^6</Tex> a perfect cubes.

<Tex block>x^6=(x^2)^3</Tex>

Thus,

<Tex block>
\begin{align*}
x^6+1&=(x^2)^3+1^3\\\
&=(x^2+1)((x^2)^2-x^2 \times1+1^2)\\\
&=(x^2+1)(x^4-x^2+1)
\end{align*}
</Tex>

<Example type="example" num="5"/>

Factor the sum: <Tex>(x+1)^3+(y+1)^3</Tex>

<Example type="solution"/>

This expression is written as a sum of two perfect cubes. The two terms are composite terms.

<Tex block>
\begin{align*}
\scriptsize (x+1)^3+(y+1)^3&=\scriptsize \big( (x+1)+(y+1)\big) \big( (x+1)^2+(x+1)(y+1)+(y+1)^2 \big) \\\
&=\scriptsize \big(x+1+y+1 \big) \big( x^2+2x+1+xy+x+y+1+y^2+2y+1 \big) \\\
&=\scriptsize \big(x+y+2 \big) \big( x^2+y^2+3x+3y+xy+3 \big)
\end{align*}
</Tex>

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Learn how to factor sum and difference of two perfect cubes using a very simple patterns. This lesson  is illustrated by many examples with solutions.

Explorer all lessons of Intermediate Algebra