# Difference of Squares – Examples with Answers

The difference of squares allows us to factor algebraic expressions. The difference of squares tells us that it is possible to write a quadratic expression as the product of two binomials, one containing the sum of the square roots and the other containing the subtraction of the square roots.

In this article, we will look at a summary of the difference of squares along with several examples with answers and exercises to solve.

##### ALGEBRA

Relevant for

Exploring several examples of differences of squares.

See examples

##### ALGEBRA

Relevant for

Exploring several examples of differences of squares.

See examples

## Summary of the difference of squares

Remember that the difference of squares is a theorem that tells us whether a quadratic equation can be written as the product of two binomials. One of these binomials shows the difference of the square roots and the other binomial shows the sum of the square roots. A difference of squares is expressed in the form:

$latex {{a}^2}-{{b}^2}$

where both the first and second terms are perfect squares. By factoring the difference of squares, we have:

$latex {{a}^2}-{{b}^2}=(a+b)(a-b)$

To factor using the difference of squares, we can follow the following steps:

Step 1: Extract the common factor if there is one. Don’t forget to include this common factor in your final answer.

Step 2: Determine the numbers that will produce the same results and use the formula $latex {{a}^2}-{{b}^2}=(a+b)(a-b)$.

Step 3: Factor and simplify the resulting expression if possible.

## Difference of squares – Examples with answers

### EXAMPLE 1

Factor the expression $latex {{x}^2}-25$.

Step 1: We have no common factors.

Step 2: Since we know that 5 squared equals 25, we can rewrite the expression as follows:

$latex {{x}^2}-{{5}^2}$

Now, we apply the formula $latex {{a}^2}-{{b}^2}=(a+b)(a-b)$:

$latex=(x+5)(x-5)$

Step 3: We can no longer factor.

### EXAMPLE 2

Use the difference of squares to factor $latex x^2-64$.

Step 1: We have no common factors.

Step 2: We can write the expression as follows:

$latex x^2-8^2$

Now, we apply the difference of squares formula $latex a^2-b^2=(a+b)(a-b)$:

$latex=(x+8)(x-8)$

Step 3: The expression is now simplified.

### EXAMPLE 3

Factor the expression $latex {{x}^2} -81$.

Step 1: The expression has no common factors.

Step 2: Since we have $latex {{x}^2}-81={{x}^2}-{{9}^2}$, we apply the formula of the difference of squares $latex {{a}^2}-{{b}^2}=(a+b)(a-b)$, where a is equal to x and b is equal to 9:

$latex=(x+9)(x-9)$

Step 3: We can no longer factor.

### EXAMPLE 4

Use the difference of squares to factor $latex 4x^2-36$.

Step 1: At first glance, it would seem that the difference of squares doesn’t apply here. However, we can extract the common factor 4 from both terms:

$latex 4x^2-36=4(x^2-9)$

Step 2: Now we can write 9 as $latex 3^2$ and apply the difference of squares formula:

$latex 4(x^2-3^2)=4(x+3)(x-3)$

Step 3: The expression is now simplified.

### EXAMPLE 5

Factor the expression $latex 4x^2-121y^2$.

Step 1: The terms have no common factors.

Step 2: We can write the terms of the expression as follows:

$latex (2x)^2-(11y)^2$

Now, we can apply the difference of squares formula, and we have:

$latex (2x)^2-(11y)^2=(2x+11y)(2x-11y)$

Step 3: The expression is now simplified.

### EXAMPLE 6

Use the difference of squares to factor $latex 8a^2-50b^2$.

Step 1: We can extract the common factor 2 from both terms:

$latex 2(4a^2-25b^2)$

Step 2: Now, we write the expression as follows:

$latex =2((2a)^2-(5b)^2)$

Using the difference of squares formula, we have:

$latex 2((2a)^2-(5b)^2)=2(2a+5b)(2a-5b)$

Step 3: The expression is now simplified.

### EXAMPLE 7

Factor the expression $latex 3{{x}^2}-27{{y}^2}$.

Step 1: In this case, 3 is a common factor of both terms, so we factor it:

$latex 3{{x}^2}-27{{y}^2}=3({{x}^2}-9{{y}^2})$

Step 2: We can write $latex 9{{y}^2}$ like $latex {{(3y)}^2}$. Now, we apply the formula for the difference of squares:

$latex 3({{x}^2}-{{(3y)}^2})=3(x+3y)(x-3y)$

Step 3: We can no longer factor.

### EXAMPLE 8

Factor the expression $latex {{x}^3}-64x$.

Step 1: Since x is a common factor, we factor it:

$latex {{x}^3}-64x=x({{x}^2}-64)$

Step 2: We can write the 64 like $latex {{8}^2}$. Using the formula for the difference of squares, we have:

$latex x({{x}^2}-64)=x(x+8)(x-8)$

Step 3: We can no longer factor.

### EXAMPLE 9

Factor the expression $latex {{(y-3)}^2}-{{(y+5)}^2}$.

Step 1: We have nothing to factor.

Step 2: In this case, we have that a is equal to $latex (y-3)$ and b is equal to $latex (y+5)$. Thus, we use the formula for the difference of squares:

$$[(y-3)+(y+5)][(y-3)-(y-5)]$$

Step 3: Here, we can combine like terms and simplify:

$latex =[2y+2]$

$latex =4y+4$

### EXAMPLE 10

Factor the expression $latex 16{{(x+y)}^2}-25{{(x-2y)}^2}$.

Step 1: We have nothing to factor.

Step 2: We start by writing the expression as follows:

$latex {{(4(x+y))}^2}-{{(5(x-2y))}^2}$

Here, we have that a is equal to $latex 4(x+y)$ and b is equal to $latex 5(x-2y)$. So, we use the difference of squares formula:

$$[4(x+y)+5(x-2y)][4(x+y)-5(x-2y)]$$

Step 3: Here we can expand, combine like terms, and simplify:

$$=[4x+4y+5x-10y][4x+4y-5x+10y]$$

$latex =(9x-6y)(-x+14y)$

## Difference of squares – Practice problems

Difference of squares quiz  You have completed the quiz!

#### Factor the expression $latex x^3-144x$ using the difference of squares.

Write the answer in the input box.

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Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  