6.7 Factor Binomials – Difference of Squares

Pooja Gupta

6.7 Factor Binomials – Difference of Squares

Learning Objectives

By the end of this section, you will be able to:

Factor Binomials of the Form a²x² – b²y²

Let us first review briefly what we learned in Chapter 6.3:

Product of Conjugates Pattern

If $a$ and $b$ are real numbers,

The product is called a difference of squares.

Factor the Difference of Squares

To be more specific, we are going to see how we can factorize the difference of two squares. From above, it is very clear that factors of a binomial of the type ${a}^{2}-{b}^{2}$ are always $\left(a+b\right)\left(a-b\right)$ . Once we know what $a$ and $b$ are, the factorization becomes very easy.

EXAMPLE 1

Factor the difference of square: ${x}^{2}-{64}$

Solution

First, we rewrite each term of ${x}^{2}-{64}$ as a perfect square of an expression.

Rewrite each term as a perfect square
Treating $x$ as $a$ and $8$ as $b$
Apply the difference of squares formula
Hence,	${x}^{2}-{64}$ $=(x+8)(x-8)$

Difference of Squares

If $a$ and $b$ are real numbers, then

difference of squares formula

TRY IT 1.1

Factor: $n^2 - 49$

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$(n+7)(n-7)$

TRY IT 1.2

Factor: ${w}^{2}-9$ .

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$\left(w+3\right)\left(w-3\right)$

EXAMPLE 2

Factor completely: $4{m}^{2}-81$ .

Solution

We know that $4m^2 = (2m)^2$ and $81 = 9^2$

Rewrite each term as a perfect square	$(2m)^2 - 9^2$
Treating $2m$ as $a$ and $9$ as $b$	$a = 2m , b = 9$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(2m+9)(2m-9)$
Hence,	$4{m}^{2}-81$ $=(2m+9)(2m-9)$

TRY IT 2.1

Factor completely: $36{x}^{2}-49$ .

Show answer

$\left(6x+7\right)\left(6x-7\right)$

TRY IT 2.2

Factor completely: $49{x}^{2}-\dfrac{25}{121}$ .

Show answer

$\left(7x+\dfrac{5}{11}\right)\left(7x-\dfrac{5}{11}\right)$

Now we’ll factor a similar binomial that has two variables.

EXAMPLE 3

Factor completely: $25{x}^{2}-{y}^2$ .

Solution

We know that $25=5^2$ , $x^2 = (x)^2$ and $y^2 = (y)^2$

Rewrite each term as a perfect square	$(5x)^2 - (y)^2$
Assigning values to $a$ and $b$	$a = 5x , b = y$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(5x+y)(5x-y)$
Hence,	$25x^2 - y^2 = (5x+y)(5x-y)$

TRY IT 3.1

Factor: $49{y}^2-16{x}^{2}$ .

Show answer

$\left(7y+4x\right)\left(7y-4x\right)$

TRY IT 3.2

Factor: $81{m}^2-4{n}^{2}$ .

Show answer

$\left(9m+2n\right)\left(9m-2n\right)$

EXAMPLE 4

Factor completely: $25{x}^{2}{y}^2 - 36$ .

Solution

We know that $25=5^2$ , $x^2 = (x)^2$ , $y^2 = (y)^2$ and $36 = 6^2$

Rewrite each term as a perfect square	$(5xy)^2 - (6)^2$
Assigning values to $a$ and $b$	$a = 5xy , b = 6$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(5xy+6)(5xy-6)$
Hence,	$25x^2y^2 - 36 = (5xy+6)(5xy-6)$

TRY IT 4.1

Factor: ${x}^{2}{y}^{2}-1$ .

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$\left(xy-1\right)\left(xy+1\right)$

TRY IT 4.2

Factor: ${a}^{2}{b}^{2}-81$ .

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$\left(ab-9\right)\left(ab+9\right)$

Let us now try to factor a binomial where it is not apparent that we can use the Difference of Squares formula right away.

EXAMPLE 5

Factor: $3a^2 - 27b^2$

Solution

We know that $a^2 = (a)^2$ and $b^2 = (b)^2$ but neither 3 nor 27 is a perfect square. In a question like this check whether the numbers have a GCF other than 1. In this case, the GCF of 3 and 27 is 3. Now apply your knowledge of factoring out the GCF and check if you can factor the binomial any further:

Factor out the GCF	$3(a^2 -9b^2)$
Rewrite each term as a perfect square	$3((a)^2 - (3b)^2)$
Assigning values to $a$ and $b$	$a = a , b = 3b$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$3(a+3b)(a-3b)$
Hence,	$3a^2 - 27b^2 = 3(a+3b)(a-3b)$

TRY IT 5.1

Factor completely: $36x^2-9y^2$

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$9(2x+y)(2x-y)$

TRY IT 5.2

Factor completely: $3.6x^2-0.9$

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$0.9(2x+1)(2x-1)$

EXAMPLE 6

Factor: $x^4 - y^6$

Solution

We know that $x^4 = (x^2)^2$ and $y^6 = (y^3)^2$

Factor out the GCF (if any)
Rewrite each term as a perfect square	$((x^2)^2 - (y^3)^2)$
Assigning values to $a$ and $b$	$a = x^2 , b = y^3$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(x^2+y^3)(x^2-y^3)$
Hence,	$x^4 - y^6 = (x^2+y^3)(x^2-y^3)$

TRY IT 6.1

Factor completely: $x^4 - 9$

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$(x^2+3)(x^2-3)$

TRY IT 6.2

Factor completely: $2m^4n^2 - 50$

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$2(m^2n+5)(m^2n-5)$

Can you factor Sum of Squares?

Note that, except for a common factor, a sum of squares or a binomial of the form $a^2 + b^2$ is not factorable over the set of real numbers.

EXAMPLE 7

Factor completely: $x^4 - 16$

Solution

We know that $x^4 = (x^2)^2$ and $16 = (4)^2$

Factor out the GCF (if any)
Rewrite each term as a perfect square	$((x^2)^2 - (4)^2)$
Assigning values to $a$ and $b$	$a = x^2 , b =4$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(x^2+4)(x^2-4)$
So far, this question is not much different from the previous examples but note that this factorization is incomplete. Why? Because the second factor $(x^2-4)$ can also be rewritten as a difference of squares. The first factor $(x^2+4)$ is a sum of squares and cannot be factored further over the set of real numbers.
Rewrite each term of the second factor as a perfect square	$(x^2+4)((x)^2 - (2)^2)$
Assigning values to $a$ and $b$	$a = x , b =2$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(x^2+4)(x+2)(x-2)$
Hence,	$x^4 - 16 = (x^2+4)(x+2)(x-2)$

TRY IT 7.1

Factor completely: $y^4 - 81$

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$(y^2 +9)(y+3)(y-3)$

TRY IT 7.2

Factor completely: $5p^4 - 80$

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$5(p^2 + 4)(p+2)(p-2)$

EXAMPLE 8

Factor completely: $36-(x-5)^2$

Solution

We can apply the difference of squares formula not only on monomials that are perfect squares but also on any polynomial that is a perfect square. In this example, we know that $36 = (6)^2$ and the second term is already in the form of a perfect square.

Factor out the GCF (if any)
Rewrite each term as a perfect square	$((6)^2 - (x-5)^2)$
Assigning values to $a$ and $b$	$a = 6 , b =x-5$
Apply the difference of squares formula $a^2 -b^2 = (a+b)(a-b)$	$(6+(x-5))(6-(x-5))$
Note the brackets around the $(x-5)$ . This is especially important in the second factor where we have a negative sign to take care of.
Rewrite to remove the inner brackets	$(6+x-5)(6-x+5)$
Combine like terms	$(1+x)(11-x)$
Hence,	$36-(x-5)^2 = (1+x)(11-x)$

TRY IT 8.1

Factor completely: $(x-7)^2 - 49$

Show answer

$x(x-14)$

TRY IT 8.2

Factor completely: $16 - (k+3)^2$

Show answer

$(k+7)(1-k)$

Practice Makes Perfect

In the following exercises, factorize using the difference of squares

1. ${c}^{2}-25$	2. $m^2 -49$
3. ${b}^{2}-\dfrac{36}{49}$	4. ${x}^2 - \dfrac{9}{16}$
5. $64{j}^{2}-16$	6. $25k^2 - 36$
7. $81{c}^{2}-25$	8. $121 k^2 - 16$
9. $169-{q}^{2}$	10. $121 - b^2$
11. $16-36{y}^{2}$	12. $25 - 9x^2$
13. $49{w}^{2}-100{x}^{2}$	14. $81c^2 - 4d^2$
15. ${p}^{2}-\dfrac{16}{25}{q}^{2}$	16. ${m}^{2}-\dfrac{4}{9}{n}^{2}$
17. ${x}^{2}{y}^{2}-81$	18. $a^2b^2 - 16$
19. ${r}^{2}{s}^{2}-\dfrac{4}{49}$	20. ${u}^{2}{v}^{2}-\dfrac{9}{25}$
21. $7x^2-28y^2$	22. $4x^2m-36y^2m$
23. $2x^4r - 72y^4r$	24. $7x^4 - 343y^4$
25. $25x^6-y^6$	26. $16m^4-n^4$
27. $16-(x-2)^2$	28. $9y^2 - (y+9)^2$
29. $0.25a^2-0.16b^2$	30. $25 - (n+3)^2$

Answers

1. $\left(c-5\right)\left(c+5\right)$	2. $\left(m-7\right)\left(m+7\right)$
3. $\left(b+\dfrac{6}{7}\right)\left(b-\dfrac{6}{7}\right)$	4. $\left(x+\dfrac{3}{4}\right)\left(x-\dfrac{3}{4}\right)$
5. $\left(8j+4\right)\left(8j-4\right)$	6. $\left(5k+6\right)\left(5k-6\right)$
7. $\left(9c+5\right)\left(9c-5\right)$	8. $\left(11k+4\right)\left(11k-4\right)$
9. $\left(13-q\right)\left(13+q\right)$	10. $\left(11-b\right)\left(11+b\right)$
11. $\left(4-6y\right)\left(4+6y\right)$	12. $\left(5-3x\right)\left(5+3x\right)$
13. $\left(7w+10x\right)\left(7w-10x\right)$	14. $\left(9c-2d\right)\left(9c+2d\right)$
15. $\left(p+\dfrac{4}{5}q\right)\left(p-\dfrac{4}{5}q\right)$	16. $\left(m+\dfrac{2}{3}n\right)\left(m-\dfrac{2}{3}n\right)$
17. $\left(xy-9\right)\left(xy+9\right)$	18. $\left(ab-4\right)\left(ab+4\right)$
19. $\left(rs-\dfrac{2}{7}\right)\left(rs+\dfrac{2}{7}\right)$	20. $\left(uv-\dfrac{3}{5}\right)\left(uv+\dfrac{3}{5}\right)$
21. $7(x+2y)(x-2y)$	22. $4m(x+3y)(x-3y)$
23. $2r(x^2 + 6y^2)(x^2 - 6y^2)$	24. $7(x^2+7y^2)(x^2-7y^2)$
25. $(5x^3 + y^3)(5x^3 - y^3)$	26. $(4m^2+n^2)(2m+n)(2m-n)$
27. $(x+2)(6-x)$	28. $(4y+9)(2y-9)$
29. $0.01(5a+4b)(5a-4b)$ OR $(0.5a+0.4b)(0.5a-0.4b)$	30. $(n+8)(2-n)$

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Factor the Difference of Squares

Practice Makes Perfect

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