Difference of Squares

The difference of squares rule is an essential tool kit to learn and understand while learning how to factor and simplify different quadratic expressions. This relationship highlights how the difference between two squares can be rewritten as a factor of two binomials. What’s more interesting is that the two factors are simply the sum and difference of the two terms’ square root. 

The difference of squares rule opens a wide range of applications as well, so it’s helpful to familiarize yourself with this rule. Even better, know this rule by heart by knowing how to derive it. Don’t worry, this article covers everything you need to know about the difference of squares rule for you!

What Is the Difference of Squares?

The difference of squares is an algebraic property that highlights the relationship shared between the difference between the squares of two terms. The difference of squares highlights that when given a difference of squares, the expression can be rewritten as a product of the terms’ sum and the terms’ difference

\begin{aligned}a^2 - b^2 = (a - b)(a + b)\end{aligned}

This equations shows that given two terms, a and b, the difference of their squares, a^2 –b^2 is simply equal to the product of their sum and difference, (a – b)(a + b). To appreciate the difference of squares rule even more, take a look at how these three expressions are factored by using this rule.

\begin{aligned}(2a)^2 - (3b)^2 &= (2a - 3b)(2a + 3b)\end{aligned} \begin{aligned}200^2 - 1^2 &= (200 -1)(200 + 1)\end{aligned}\begin{aligned}4x^2 - 9y^2 &= (2x)^2 - (3y)^2\\&= (2x -3y)(2x + 3y)\end{aligned}

These three examples highlight an important takeaway: when given a difference of two squares, the expression can still be factored using its terms’ sum and difference. Before learning how to apply this rule, it’s important that you know how to easily spot for expressions that will benefit from this rule.

Problem 1

Determine whether the following expressions will benefit from the difference of squares rule. 

a. 99^2 + 1^2

99^2 {\color{Orchid}+} 1^2

By taking a look at the operation (addition) between the two terms, you can see that the difference of squares rule won’t work for this expression.  

b. 25m^2 – 9n^2

Now, since the two terms are being subtracted, continue expecting whether the two terms can be expressed as two squares. 

\begin{aligned}25m^2 &= (5m)^2\\9n^2 &= (3n)^2\\25m^2 - 9n^2 &= (5m)^2 - (3n)^2\end{aligned}

Since the two terms are both perfect squares, the difference of squares rule can be applied to factor the given expression.

c. 21x^2 - y

Although the two terms are being subtracted with the leading term having a quadratic term, the second term only has a power of 1. 

21x^{\color{Orange}2} - y^{{\color{Orange}1}}

The two terms have to be perfect squares for the difference of squares rule to apply, so it won’t work for this item as well. 

Identifying when the difference of squares comes in handy is just the start. But before learning how to apply the rule, why don’t you take a look at the difference of squares rule’s proof. Not interested? No worries, it’s just a complementary section for the curious ones. Feel free to move to the next section if you’re excited to use this rule right away!

Understanding the Proof of Difference of Squares

To understand why the difference of squares rule applies, use the rule for expanding binomials and work on the right-hand side of the difference of squares rule.

\begin{aligned}a^2 - b^2 = {\color{Teal}(a - b)(a + b)}\end{aligned}

Apply the FOIL method to expand (a –b)(a+b). Simplify the expression by combining like terms as shown below.

\begin{aligned}(a -b)(a + b) &= a(a) -b(a) + a(b) -b(b)\\&=a^2 -\cancel{ab}+ \cancel{ab}-b^2\\&=a^2-b^2\end{aligned}

After simplifying the right-hand side of the difference of squares rule, it you can see that the resulting expression is equivalent to the left-hand side. 

\begin{aligned}a^2 -b^2 &= (a -b)(a+ b)\\&\overset{\checkmark}{=} a^2 -b^2\end{aligned}

Thus, the difference of squares rule is true for all real numbers and algebraic expressions. Now that this has been established, it’s time to use the rule to evaluate and factor expressions. 

How to Apply the Difference of Squares?

When applying the difference of squares rule or formula, the first step is of course, is to identify whether the two terms involved are each perfect squares. Once this has been confirmed apply the following steps to utilize the difference of squares rule correctly:

  1. Express each term as a perfect square if it’s still not. 
  2. Write down the first and last term without the perfect square. 
  3. Find the sum and the difference between the two terms. 
  4. Write the product of these two – depending on the form that is needed. 

Oftentimes, you use the difference of squares rule to factor expressions of the form, a^2 –b^2. You’ll sometimes encounter problems that involve evaluating complex arithmetic expressions using this rule. Don’t worry, this section covers both for you!

Factoring Expressions Using Difference of Squares

When factoring expressions using the difference of square rule, a^2 –b^2 = (a- b)(a +b), the first step is to identify the expression for a and b. Once you have this, write the difference of squares as a product of the following: 1) sum of a and b and 2) difference of a and b

Here are some examples of expressions that can be factored using the difference of squares rule:

\begin{aligned}x^2 - y^2 &= (x)^2 - (y)^2\\&= (x -y)(x + y)\end{aligned}
\begin{aligned}81m^2 - 64n^2&= (9m)^2 - (8n)^2\\&= (9m -8n)(9m + 8n)\end{aligned}
\begin{aligned}x^4 - y^4 &= (x^2)^2 -(y^2)^2\\&= (x^2 - y^2)(x^2 + y^2)\\&= (x - y)(x +y)(x^2 +y^2)\end{aligned}

The three examples show how important it is to immediately identify the perfect squares and their corresponding square root. For the second expression, 81m^2 – 64n^2, it’s important to immediately establish that 81m^2 = (9m)^2 and 64n^2 = (8n)^2. Once you’ve identified the square roots, use this to factor the expression. Hence, 81m^2 -64n^2= (9m -8n)(9m + 8n).

\begin{aligned}x^4 - y^4 &= (x^2)^2 -(y^2)^2\\&= (x^2 - y^2)(x^2 + y^2)\end{aligned}

Now, take a look at the third example, x^4 – y^4. This part shows how the expression is initially factored after rewriting x^4 and y^4 as the squares of x^2 and y^2, respectively. But the expression is not yet factored completely since x^2 – y^2 can still be factored. Hence, we have its final form as shown:

\begin{aligned} (x^2 - y^2)(x^2 + y^2)\\&= (x - y)(x +y)(x^2 +y^2)\end{aligned}

That’s an important reminder for you: double-check and see if the expression is already factored completely. Don’t worry, you’ll get to try factoring more expressions like these three in the practice problem that follows!

Problem 2

Test your understanding by factoring the following expressions completely. 

a. m^2 – n^2

You can see that the two terms are perfect square, so the difference of squares rule applies. Factor the expression by rewriting it as a product of the sum and difference of m and n as shown below. 

\begin{aligned}{\color{DarkBlue}m^2} -{\color{DarkRed}n^2} &= {\color{DarkBlue}(m)^2} -{\color{DarkRed}(n)^2} \\&= ({\color{DarkBlue}m}+{\color{DarkRed}n})({\color{DarkBlue}m} -{\color{DarkRed}n})\end{aligned}

Hence, the factored form of m^2 – n^2 is (m –n)(m + n).

b. 25a^2 – 36b^2

Whenever you see coefficients when applying the difference of squares, immediately think of the coefficients’ square roots as well. For example, you have 25 = 52 and 36 = 62. This way, once you express each term as a perfect square, you immediately know the corresponding coefficients of terms’ square roots. 

\begin{aligned}{\color{DarkGreen}25a^2} -{\color{DarkOrange}36b^2} &= {\color{DarkGreen}(5a)^2} -{\color{DarkOrange}(6b)^2} \\&= ({\color{DarkGreen}5a}+{\color{DarkOrange}6b})({\color{DarkGreen}5a}-{\color{DarkOrange}6b})\end{aligned}

From this, we have 25a^2 – 36b^2 = (5a + 6b)(5a -6b) as the fully factored form. 

c. 16p^4 – 81q^4

Now, apply a similar process to factor the binomial, 16p^4 -9q^4. Begin by writing each term as a perfect square – keeping in mind that p^4 = (p^2)^2 and q^4 = (q^2)^2.

\begin{aligned}{\color{Orchid}16p^4} -{\color{Teal}81q^4} &= {\color{Orchid}(4p^2)^2} -{\color{Teal}(9q^2)^2} \\&= ({\color{Orchid}4p^2}+{\color{Teal}9q^2})({\color{Orchid}4p^2}-{\color{Teal}9q^2})\end{aligned}

Now, inspect the second factor, (4p^2 – 9q^2). Each term is also a perfect square, so this binomial can still be factored even further using the difference of squares rule.

\begin{aligned}16p^4 - 81q^2 &= (4p^2 + 9q^2)({\color{Orchid}4p^2}-{\color{Teal}9q^2})\\&= (4p^2 + 9q^2)[{\color{Orchid}(2p)^2}-{\color{Teal}(3q)^2}]\\&= (4p^2 + 9q^2)({\color{Orchid}2p}+{\color{Teal}3q})({\color{Orchid}2p} -{\color{Teal}3q})\end{aligned}

This shows that when factored completely, 16p^4 – 81q^4 , is equivalent to (4p^2 + 9q^2)(2p + 3q)(2p -3q).

Evaluating Expressions Using Difference of Squares

When evaluating numerical expressions using difference of squares, the components to look out for are the following:

1. The two terms are perfect squares or you see two factors that are two numbres’ sum and difference.

2. See if either one simplifies to a multiple of 10, 100, 1000, or any powers of 10. 

When they meet these conditions, check if the difference of squares rules applies. For example, when working with 88^2 – 12^2, you know that the sum of these two terms is equal to 100. Since the expression is written as a difference of two terms (882 and 122). Why don’t you try applying the difference squares rules? 

\begin{aligned}88^2 - 12^2 &= (88 - 12)(88 + 12)\\&= 64(100)\\&= 6400\end{aligned}

This example shows that by applying the difference of squares rule, evaluating the expression, 88^2 – 12^2, becomes much more simpler. You didn’t have to evaluate each of the perfect square first, but instead, reduced the expression so that one becomes a multiple of 100. 

Use a similar process to evaluate the numerical expressions shown below using the difference of squares rule.

Problem 3

Evaluate the following numerical expressions. 

a. 64^2 – 36^2

Similar to the example, since 642 and 362 are both perfect squares, the difference of squares rule can be applied. Take a look at what happens to the factor when written using the rule:

\begin{aligned}64^2 - 36^2 &= (64 + 36)(64 - 36)\\&=(100)(28)\\&= 2800\end{aligned}

Hence, 64^2 – 36^2 is equivalent to 2800.

b. 499 \times 501

Now, this problem is a bit trickier. Before applying any formula, why don’t you rewrite the expression knowing that 500 = 500 - 1 and 501 = 500 + 1.

499 \times 501 = (500 -1)(500 + 1)

Now, does this factored form look familiar? It’s the right-hand side of the difference of squares rule. This means that (500 -1)(500 + 1) is simply equal to the difference of the squares of 500 and 1. Rewrite this expression in this form to evaluate it much faster. 

\begin{aligned}(500 -1)(500 + 1) &= 500^2 - 1^2\\&= 250000 - 1\\&= 24999\end{aligned}

Hence, through the difference of squares rules, we’ve simplified the process of evaluating 499 \times 501. In fact, it’s equal to 24999.