To factor a difference of squares, we need to start by applying a square root to both terms of the expression given. Then, we write the algebraic expression as a product of the sum of the terms and a difference of the terms. **The difference of squares theorem tells us that if we have an expression of the form a²-b², this is equivalent to (a+b)(a–b).**

In this article, we will look at the difference of squares in more detail. We will look at how to factor the difference of squares by using a formula, and we will look at worked-out examples to understand the concepts.

## Definition of difference of squares

The difference of two squares is a theorem that tells us if a quadratic equation can be written as a product of two binomials, where one shows the difference of the square roots and the other shows the sum of the square roots.

A difference of squares is something that looks like $latex {{x}^2} – 4$. This is because $latex {{2} ^ 2} = 4$, so we actually have $latex {{x} ^ 2}- {{2} ^ 2}$, which is a difference of squares.

## Difference of squares formula

The difference of squares formula is an algebraic expression that is used to express the difference between two squared values. A difference of squares is expressed in the form:

$latex {{a}^2} – {{b}^2}$

where the first and last terms are perfect squares.

Factoring the difference of squares, we have:

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

This is true because $$(a+b)(a-b)={{a}^{2}}+ab-ab-{{b}^{2}}={{a}^{2}}-{{b}^{2}}$$

## How to factor a difference of squares

The following are the steps required to factor a difference of squares:

** Step 1:** Determine if the terms have something in common, called the greatest common factor. If so, factor out that common factor, and don’t forget to include it in your final answer.

**Step 2:** All difference of squares problems can be factored as follows: $latex {{a}^2}-{{b}^2} = (a + b) (a-b)$. Therefore, all we have to do is to find the square numbers that will produce the desired results.

**Step 3:** Determine if the remaining factors can be factored further.

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## Difference of squares – Examples with answers

The following examples follow the process outlined above to factor the difference of squares:

### EXAMPLE 1

Factor $latex {{x}^{2}}-9$.

**Step 1:** Determine if the terms have anything in common: they have nothing in common.

**Step 2: **To factor the problem in the form $latex (a+b)(a-b)$, we need to determine the value that we have to square to get $latex {{x}^{2}}$ and the value that we have to square to obtain $latex 9$. In this case, we have

*x*and 3 because $latex (x)(x)={{x}^{2}}$ and $latex (3)(3)=9$.

$latex (x+3)(x-3)$

**Step 3: **Determine whether the remaining factors can be factored: in this case no.

### EXAMPLE 2

Factor $latex 4{{x}^{2}}-49$.

**Step 1:** Determine if the terms have anything in common: they have nothing in common.

**Step 2: **To factor the problem in the form $latex (a+b)(a-b)$, we need to determine the value that we have to square to get $latex 4{{x}^{2}}$ and the value that we have to square to obtain $latex 49$. In this case, we have 2

*x*and 7 because $latex (2x)(2x)=4{{x}^{2}}$ and $latex (7)(7)=49$.

$latex (2x+7)(2x-7)$

**Step 3: **Determine whether the remaining factors can be factored: in this case no.

### EXAMPLE 3

Factor $latex 18{{x}^{2}}-98$.

**Step 1:** Determine if the terms have anything in common: they have 2 common.

$latex 2(9{{x}^2}-49)$

**Step 2: **To factor the problem in the form $latex (a+b)(a-b)$, we need to determine the value that we have to square to get $latex 9{{x}^{2}}$ and the value that we have to square to obtain $latex 49$. In this case, we have 3

*x*and 7 because $latex (3x)(3x)=9{{x}^{2}}$ and $latex (7)(7)=49$.

$latex 2(3x+7)(3x-7)$

**Step 3: **Determine whether the remaining factors can be factored: in this case no.

### EXAMPLE 4

Factor $latex 4{{x}^{2}}-64$.

**Step 1:** Determine if the terms have anything in common: they have 4 common.

$latex 4({{x}^2}-16)$

**Step 2: **To factor the problem in the form $latex (a+b)(a-b)$, we need to determine the value that we have to square to get $latex {{x}^{2}}$ and the value that we have to square to obtain $latex 16$. In this case, we have

*x*and 4 because $latex (x)(x)={{x}^{2}}$ and $latex (4)(4)=16$.

$latex 4(x+4)(x-4)$

**Step 3: **Determine whether the remaining factors can be factored: in this case no.

### EXAMPLE 5

Factor $latex 16{{x}^{4}}-1$.

**Step 1:** Determine if the terms have anything in common: they have nothing in common.

**Step 2: **To factor the problem in the form $latex (a+b)(a-b)$, we need to determine the value that we have to square to get $latex 16{{x}^{4}}$ and the value that we have to square to obtain $latex 1$. In this case, we have $latex 4{{x}^2}$ and 1 because $latex (4{{x}^2})(4{{x}^2})=16{{x}^{4}}$ and $latex (1)(1)=1$.

$latex (4{{x}^2}+1)(4{{x}^2}-1)$

**Step 3: **Determine whether the remaining factors can be factored: in this case one of the factors is a difference of squares, so we can factor it: we need to determine the value we have to square to get $latex 4{{x}^2}$ and the value we have to square to obtain 1. In this case, we have $latex 2x$ and 1 since $latex (2x) (2x) = 4 {{x} ^2}$ and $latex (1)(1) = 1$.

$latex (4{{x}^{2}}+1)(2x+1)(2x-1)$

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## Difference of squares – Practice problems

## See also

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