1. Learning objectives

Learning objectives

Factor perfect square trinomials

Factor differences of squares

Factor sums and differences of cubes

We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

2. Factor Perfect Square Trinomials

Factor Perfect Square Trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.

The trinomial $9{x}^{2}+24x+16$ is called a perfect square trinomial. It is the square of the binomial $3x+4.$

In this chapter, you will start with a perfect square trinomial and factor it into its prime factors.

You could factor this trinomial using the methods described in the last section, since it is of the form $a{x}^{2}+bx+c.$ But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work.

Here is the pattern—the reverse of the binomial squares pattern.

If a and b are real numbers

\begin{array}{c} {a}^{2}+2ab+{b}^{2}={(a+b)}^{2} \\ {a}^{2}-2ab+{b}^{2}={(a-b)}^{2} \\ \end{array}

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, ${a}^{2}.$ Next check that the last term is a perfect square, ${b}^{2}.$ Then check the middle term—is it the product, $2ab?$ If everything checks, you can easily write the factors.

3. Example 6.23

Example 6.23

How to Factor Perfect Square Trinomials

Factor: $9{x}^{2}+12x+4.$

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern ${a}^{2}-2ab+{b}^{2},$ which factors to ${(a-b)}^{2}.$

The steps are summarized here.

Factor perfect square trinomials.

$\begin{array}{llcccccc} \mathrm{Step 1.} & \mathrm{Does the trinomial fit the pattern?} & & & \,{a}^{2}+2ab+{b}^{2} & & & {a}^{2}-2ab+{b}^{2} \\ & \mathrm{Is the first term a perfect square?} & & & \,{(a)}^{2} & & & {(a)}^{2} \\ & \mathrm{Write it as a square.} & & & & & & \\ & \mathrm{Is the last term a perfect square?} & & & \,{(a)}^{2}\,{(b)}^{2} & & & {(a)}^{2}\,{(b)}^{2} \\ & \mathrm{Write it as a square.} & & & & & & \\ & \mathrm{Check the middle term. Is it}\,2ab? & & & \,{(a)}^{2}{}_{↘}\underset{2\cdot{}a\cdot{}b}{}{}_{↙}{(b)}^{2} & & & {(a)}^{2}{}_{↘}\underset{2\cdot{}a\cdot{}b}{}{}_{↙}{(b)}^{2} \\ \mathrm{Step 2.} & \mathrm{Write the square of the binomial.} & & & \,{(a+b)}^{2} & & & {(a-b)}^{2} \\ \mathrm{Step 3.} & \mathrm{Check by multiplying.} & & & & & & \\ \end{array}$

We’ll work one now where the middle term is negative.

4. Example 6.24

Example 6.24

Factor: $81{y}^{2}-72y+16.$

Solution

The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be ${(a-b)}^{2}.$

The next example will be a perfect square trinomial with two variables.

5. Example 6.25

Example 6.25

Factor: $36{x}^{2}+84xy+49{y}^{2}.$

Solution

Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

6. Example 6.26

Example 6.26

Factor: $100{x}^{2}y-80xy+16y.$

Solution

Remember: Keep the factor 4*y* in the final product.

Check: $\,\begin{array}{c} 4y{(5x-2)}^{2} \\ 4y[{(5x)}^{2}-2\cdot{}5x\cdot{}2+{2}^{2}] \\ 4y(25{x}^{2}-20x+4) \\ 100{x}^{2}y-80xy+16y\mathrm{✓} \\ \end{array}$

7. Factor Differences of Squares

Factor Differences of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

A difference of squares factors to a product of conjugates.

If a and b are real numbers,

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

8. Example 6.27

Example 6.27

How to Factor a Binomial Using the Difference of Squares

Factor: $64{y}^{2}-1.$

Factor differences of squares.

$\begin{array}{llccc} \mathrm{Step 1.} & \mathrm{Does the binomial fit the pattern?} & & & \,{a}^{2}-{b}^{2} \\ & \mathrm{Is this a difference?} & & & \,\underline{\hspace{2em}}-\underline{\hspace{2em}} \\ & \mathrm{Are the first and last terms perfect squares?} & & & \\ \mathrm{Step 2.} & \mathrm{Write them as squares.} & & & \,{(a)}^{2}-{(b)}^{2} \\ \mathrm{Step 3.} & \mathrm{Write the product of conjugates.} & & & \,(a-b)(a+b) \\ \mathrm{Step 4.} & \mathrm{Check by multiplying.} & & & \\ \end{array}$

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression ${a}^{2}+{b}^{2}$ is prime!

The next example shows variables in both terms.

9. Example 6.28

Example 6.28

Factor: $144{x}^{2}-49{y}^{2}.$

Solution

$144{x}^{2}-49{y}^{2}$

Is this a difference of squares? Yes. · ${(12x)}^{2}-{(7y)}^{2}$ Factor as the product of conjugates. · $(12x-7y)(12x+7y)$

Check by multiplying. $\,\begin{array}{c} (12x-7y)(12x+7y) \\ 144{x}^{2}-49{y}^{2}\mathrm{✓} \\ \end{array}$

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

10. Example 6.29

Example 6.29

Factor: $48{x}^{4}{y}^{2}-243{y}^{2}.$

Solution

$48{x}^{4}{y}^{2}-243{y}^{2}$

Is there a GCF? Yes, $3{y}^{2}$ —factor it out! · $3{y}^{2}(16{x}^{4}-81)$ Is the binomial a difference of squares? Yes. · $3{y}^{2}({(4{x}^{2})}^{2}-{(9)}^{2})$ Factor as a product of conjugates. · $3{y}^{2}(4{x}^{2}-9)(4{x}^{2}+9)$ Notice the first binomial is also a difference of squares! · $3{y}^{2}({(2x)}^{2}-{(3)}^{2})(4{x}^{2}+9)$ Factor it as the product of conjugates. · $3{y}^{2}(2x-3)(2x+3)(4{x}^{2}+9)$

The last factor, the sum of squares, cannot be factored.

Check by multiplying:

$3{y}^{2}(2x-3)(2x+3)(4{x}^{2}+9)$

$3{y}^{2}(4{x}^{2}-9)(4{x}^{2}+9)$

$3{y}^{2}(16{x}^{4}-81)$

$48{x}^{4}{y}^{2}-243{y}^{2}\mathrm{✓}$

The next example has a polynomial with 4 terms. So far, when this occurred we grouped the terms in twos and factored from there. Here we will notice that the first three terms form a perfect square trinomial.

11. Example 6.30

Example 6.30

Factor: ${x}^{2}-6x+9-{y}^{2}.$

Solution

Notice that the first three terms form a perfect square trinomial.

You may want to rewrite the solution as $(x-y-3)(x+y-3).$

12. Factor Sums and Differences of Cubes

Factor Sums and Differences of Cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

\begin{array}{c} {a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2}) \\ {a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2}) \\ \end{array}

We’ll check the first pattern and leave the second to you.

\begin{array}{c} {a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2}) \\ {a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2}) \\ \end{array}

The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.

The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It will be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table 6.1.

n · 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 ${n}^{3}$ · 1 · 8 · 27 · 64 · 125 · 216 · 343 · 512 · 729 · 1000

Table

6.1

13. Example 6.31

Example 6.31

How to Factor the Sum or Difference of Cubes

Factor: ${x}^{3}+64.$

Factor the sum or difference of cubes.

Step 1. Does the binomial fit the sum or difference of cubes pattern?Is it a sum or difference?Are the first and last terms perfect cubes?
Step 2. Write them as cubes.
Step 3. Use either the sum or difference of cubes pattern.
Step 4. Simplify inside the parentheses.
Step 5. Check by multiplying the factors.

14. Example 6.32

Example 6.32

Factor: $27{u}^{3}-125{v}^{3}.$

Solution

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

15. Example 6.33

Example 6.33

Factor: $6{x}^{3}y+48{y}^{4}.$

Solution

Check:

To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by $6y.$ We’ll leave the multiplication for you.

The first term in the next example is a binomial cubed.

16. Example 6.34

Example 6.34

Factor: ${(x+5)}^{3}-64{x}^{3}.$

Solution

17. Section 6.3 Exercises

Section 6.3 Exercises

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

Mixed Practice

In the following exercises, factor completely.

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

18. Practice 1

Factor: $4{x}^{2}+12x+9.$

Factor: $9{y}^{2}+24y+16.$

Factor: $64{y}^{2}-80y+25.$

Factor: $16{z}^{2}-72z+81.$

19. Practice 2

Factor: $49{x}^{2}+84xy+36{y}^{2}.$

Factor: $64{m}^{2}+112mn+49{n}^{2}.$

Factor: $8{x}^{2}y-24xy+18y.$

Factor: $27{p}^{2}q+90pq+75q.$

20. Practice 3

Factor: $121{m}^{2}-1.$

Factor: $81{y}^{2}-1.$

Factor: $196{m}^{2}-25{n}^{2}.$

Factor: $121{p}^{2}-9{q}^{2}.$

21. Practice 4

Factor: $2{x}^{4}{y}^{2}-32{y}^{2}.$

Factor: $7{a}^{4}{c}^{2}-7{b}^{4}{c}^{2}.$

Factor: ${x}^{2}-10x+25-{y}^{2}.$

Factor: ${x}^{2}+6x+9-4{y}^{2}.$

22. Practice 5

Factor: ${x}^{3}+27.$

Factor: ${y}^{3}+8.$

Factor: $8{x}^{3}-27{y}^{3}.$

Factor: $1000{m}^{3}-125{n}^{3}.$

23. Practice 6

Factor: $500{p}^{3}+4{q}^{3}.$

Factor: $432{c}^{3}+686{d}^{3}.$

Factor: ${(y+1)}^{3}-27{y}^{3}.$

Factor: ${(n+3)}^{3}-125{n}^{3}.$

24. Practice 7

$16{y}^{2}+24y+9$

$36{s}^{2}+84s+49$

$100{x}^{2}-20x+1$

$25{n}^{2}-120n+144$

25. Practice 8

$49{x}^{2}+28xy+4{y}^{2}$

$100{y}^{2}-20y+1$

$10j{k}^{2}+80jk+160j$

$75{u}^{4}-30{u}^{3}v+3{u}^{2}{v}^{2}$

26. Practice 9

$25{v}^{2}-1$

$4-49{x}^{2}$

$6{p}^{2}{q}^{2}-54{p}^{2}$

$24{p}^{2}+54$

27. Practice 10

$121{x}^{2}-144{y}^{2}$

$169{c}^{2}-36{d}^{2}$

$16{z}^{4}-1$

$162{a}^{4}{b}^{2}-32{b}^{2}$

28. Practice 11

${x}^{2}-16x+64-{y}^{2}$

${a}^{2}+6a+9-9{b}^{2}$

${x}^{3}+125$

${z}^{6}-27$

29. Practice 12

$8-343{t}^{3}$

$8{y}^{3}-125{z}^{3}$

$216{a}^{3}+125{b}^{3}$

$7{k}^{3}+56$

30. Practice 13

$2{x}^{2}-16{x}^{2}{y}^{3}$

${(x+3)}^{3}+8{x}^{3}$

${(y-5)}^{3}-64{y}^{3}$

$64{a}^{2}-25$

31. Practice 14

$27{q}^{2}-3$

$16{x}^{2}-72x+81$

$8{p}^{2}+2$

$125-8{y}^{3}$

32. Practice 15

$45{n}^{2}+60n+20$

${x}^{2}-10x+25-{y}^{2}$

${(x+1)}^{3}+8{x}^{3}$

Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

33. Practice 16

Explain why ${n}^{2}+25\ne{(n+5)}^{2}.$ Use algebra, words, or pictures.