MX Algebra Factor Special Products

Section 6.3Factor Special Products

Definition

Before you get started, take this readiness quiz.

1

Simplify: \({(3{x}^{2})}^{3}.\)
If you missed this problem, review Example 7.

\(27{x}^{6}\)

Definition
2

Multiply: \({(m+4)}^{2}.\)
If you missed this problem, review Example 8.

\({m}^{2}+8m+16\)

Definition
3

Multiply: \((x-3)(x+3).\)
If you missed this problem, review Example 9.

\({x}^{2}-9\)

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.

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.

In open parentheses 3x plus 4 close parentheses squared, 3x is a and 4 is b. Writing it as a squared plus 2ab plus b squared, we get open parentheses 3x close parentheses squared plus 2 times 3x times 4 plus 4 squared. This is equal to 9 x squared plus 24x plus 16.

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.

Perfect Square Trinomials Pattern

If a and b are real numbers

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

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.

Example 1How to Factor Perfect Square Trinomials

Factor: \(9{x}^{2}+12x+4.\)

Check that the first and last terms are perfect squares and that the middle term equals 2ab.

Step 1 is to check if the trinomial fits the perfect square trinomials pattern, a squared plus 2ab plus b squared. For this we check if the first term is a perfect square. 9 x squared is the square of 3x. Next we check if the last term is a perfect square. 4 is the square of 2. Next we check if the middle term is 2ab. 12 x is twice 3x times 2. Hence we have a perfect square trinomial. Step 2 is to write this as the square of a binomial. We write it as open parentheses 3x plus 2 close parentheses squared. Step 3 is to check by multiplying.
Try It #1

Factor: \(4{x}^{2}+12x+9.\)

\({(2x+3)}^{2}\)

Did you get it?
Try It #2

Factor: \(9{y}^{2}+24y+16.\)

\({(3y+4)}^{2}\)

Did you get it?

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.

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

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

Example 2

Factor: \(81{y}^{2}-72y+16.\)

Since the middle term is negative, check whether it fits the \({(a-b)}^{2}\) pattern.

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 \({(a-b)}^{2}.\)

Table 1
A mathematical expression shows the quadratic polynomial 81y^2 - 72y + 16, which is a perfect square trinomial.
Are the first and last terms perfect squares?Two mathematical expressions are shown: (9y)^2 on the left and (4)^2 on the right.
Check the middle term.A mathematical image illustrates the expansion of a binomial squared. The expressions (9y)^2 and (4)^2 point to 2(9y)(4), which simplifies to 72y. This shows the calculation of the 2ab term.
Does it match \({(a-b)}^{2}?\) Yes.An example of a perfect square trinomial (a-b)^2 = a^2 - 2ab + b^2, specifically (9y)^2 - 2 * 9y * 4 + 4^2.
Write as the square of a binomial.The mathematical expression (9y-4)^2 is shown in black text on a white background. It represents the quantity (9y minus 4) raised to the power of 2, indicating that the binomial should be squared.
Check by multiplying:

\(\,\begin{array}{l}{(9y-4)}^{2} \\ {(9y)}^{2}-2·9y·4+{4}^{2} \\ 81{y}^{2}-72y+16✓\end{array}\)
Try It #3

Factor: \(64{y}^{2}-80y+25.\)

\({(8y-5)}^{2}\)

Did you get it?
Try It #4

Factor: \(16{z}^{2}-72z+81.\)

\({(4z-9)}^{2}\)

Did you get it?

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

Example 3

Factor: \(36{x}^{2}+84xy+49{y}^{2}.\)

Identify a and b as the square roots of the first and last terms, then verify the middle term is 2ab.

Table 2
A mathematical expression displaying the quadratic trinomial 36x^2 + 84xy + 49y^2.
Test each term to verify the pattern.The perfect square trinomial formula (a^2 + 2ab + b^2) demonstrated with an example: (6x)^2 + 2(6x)(7y) + (7y)^2.
Factor.The mathematical expression (6x + 7y) ² is centered on a white background, representing the square of a binomial.
Check by multiplying.

\(\,\begin{array}{l}{(6x+7y)}^{2} \\ {(6x)}^{2}+2·6x·7y+{(7y)}^{2} \\ 36{x}^{2}+84xy+49{y}^{2}✓\end{array}\)
Try It #5

Factor: \(49{x}^{2}+84xy+36{y}^{2}.\)

\({(7x+6y)}^{2}\)

Did you get it?
Try It #6

Factor: \(64{m}^{2}+112mn+49{n}^{2}.\)

\({(8m+7n)}^{2}\)

Did you get it?

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.

Example 4

Factor: \(100{x}^{2}y-80xy+16y.\)

Factor out the GCF first, then check whether what remains is a perfect square trinomial.

Table 3
The image displays the algebraic expression 100x^2y - 80xy + 16y. This is a trinomial with three terms, each containing variables x and y, and constant coefficients.
Is there a GCF? Yes, \(4y,\) so factor it out.The mathematical expression 4y(25x^2 - 20x + 4) is displayed.
Is this a perfect square trinomial?
Verify the pattern.A mathematical expression showing the expansion of a perfect square trinomial inside brackets, specifically 4y[(5x)^2 - 2 * 5x * 2 + 2^2], with a^2 - 2ab + b^2 annotated in red above it.
Factor.A mathematical expression showing 4y multiplied by the quantity (5x minus 2) squared, written as 4y(5x - 2)×2.

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

Check:

\(\,\begin{array}{l}4y{(5x-2)}^{2} \\ 4y[{(5x)}^{2}-2·5x·2+{2}^{2}] \\ 4y(25{x}^{2}-20x+4) \\ 100{x}^{2}y-80xy+16y✓\end{array}\)

Try It #7

Factor: \(8{x}^{2}y-24xy+18y.\)

\(2y{(2x-3)}^{2}\)

Did you get it?
Try It #8

Factor: \(27{p}^{2}q+90pq+75q.\)

\(3q{(3p+5)}^{2}\)

Did you get it?

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:

We have open parentheses 3x minus 4 close parentheses open parentheses 3x plus 4. This is of the form a minus b, a plus b. We rewrite as open parentheses 3x close parentheses squared minus 4 squared. Here, 3x is a and 4 is b. This is equal to 9 x squared minus 16.

A difference of squares factors to a product of conjugates.

Difference of Squares Pattern

If a and b are real numbers,

a squared minus b squared equals a minus b, a plus b. Here, a squared minus b squared is difference of squares and a minus b, a plus b are conjugates.

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.

Example 5How to Factor a Binomial Using the Difference of Squares

Factor: \(64{y}^{2}-1.\)

Confirm this is a difference (subtraction) of two perfect squares.

Step 1 is to check if the binomial 64 y squared minus 1 fits the pattern. For that we check the following: Is this a difference? Yes. Are the first and last terms perfect squares? Yes. Step 2 is to write both terms as squares, So, we have open parentheses 8y close parentheses squared minus 1 squared. Step 3 is to write the product of conjugates 8y minus 1, 8y plus 1. Step 4 is to check. We multiply to get the original binomial
Try It #9

Factor: \(121{m}^{2}-1.\)

\((11m-1)(11m+1)\)

Did you get it?
Try It #10

Factor: \(81{y}^{2}-1.\)

\((9y-1)(9y+1)\)

Did you get it?
Factor differences of squares.

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

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.

Example 6

Factor: \(144{x}^{2}-49{y}^{2}.\)

Write each term as a square to identify a and b.

Table 4
\(144{x}^{2}-49{y}^{2}\)
Is this a difference of squares? Yes.\({(12x)}^{2}-{(7y)}^{2}\)
Factor as the product of conjugates.\((12x-7y)(12x+7y)\)
Check by multiplying.
\(\,\begin{array}{l} \\ \\ (12x-7y)(12x+7y) \\ 144{x}^{2}-49{y}^{2}✓\end{array}\)
Try It #11

Factor: \(196{m}^{2}-25{n}^{2}.\)

\((14m-5n)(14m+5n)\)

Did you get it?
Try It #12

Factor: \(121{p}^{2}-9{q}^{2}.\)

\((11p-3q)(11p+3q)\)

Did you get it?

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!

Example 7

Factor: \(48{x}^{4}{y}^{2}-243{y}^{2}.\)

Factor out the GCF, \(3{y}^{2},\) first.

Table 5
\(48{x}^{4}{y}^{2}-243{y}^{2}\)
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:
\(3{y}^{2}(2x-3)(2x+3)(4{x}^{2}+9)\)
\(3{y}^{2}(4{x}^{2}-9)(4{x}^{2}+9)\)
\(3{y}^{2}(16{x}^{4}-81)\)
\(48{x}^{4}{y}^{2}-243{y}^{2}✓\)
Try It #13

Factor: \(2{x}^{4}{y}^{2}-32{y}^{2}.\)

\(2{y}^{2}(x-2)(x+2)({x}^{2}+4)\)

Did you get it?
Try It #14

Factor: \(7{a}^{4}{c}^{2}-7{b}^{4}{c}^{2}.\)

\(7{c}^{2}(a-b)(a+b)({a}^{2}+{b}^{2})\)

Did you get it?

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.

Example 8

Factor: \({x}^{2}-6x+9-{y}^{2}.\)

Group the first three terms — they form a perfect square trinomial.

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

Table 6
A mathematical expression displays x squared minus 6x plus 9 minus y squared.
Factor by grouping the first three terms.A mathematical expression showing x^2 - 6x + 9 - y^2, with a brace underneath the first three terms (x^2 - 6x + 9) indicating they form a group.
Use the perfect square trinomial pattern.The mathematical expression (x-3)^2 - y^2 is displayed, representing the difference of two squares with a binomial as the first squared term.
Is this a difference of squares? Yes.
Yes—write them as squares.Two mathematical expressions are displayed: a^2 - b^2 in red text, and (x-3)^2 - y^2 in black text, both representing the difference of squares.
Factor as the product of conjugates.The image shows an algebraic expression `((x-3)-y)((x-3)+y)`, which is an application of the difference of squares formula (A-B)(A+B).
A mathematical expression showing the product of two binomials: (x - 3 - y)(x - 3 + y). This is an example of the difference of squares formula.

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

Try It #15

Factor: \({x}^{2}-10x+25-{y}^{2}.\)

\((x-5-y)(x-5+y)\)

Did you get it?
Try It #16

Factor: \({x}^{2}+6x+9-4{y}^{2}.\)

\((x+3-2y)(x+3+2y)\)

Did you get it?

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.

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

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

Table 7
The algebraic identity for the sum of cubes: (a + b)(a^2 - ab + b^2).
Distribute.A mathematical expression showing the sum of two terms: a multiplied by (a squared minus ab plus b squared) plus b multiplied by (a squared minus ab plus b squared). This simplifies to a cubed plus b cubed.
Multiply.A mathematical expression: a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3. This expression simplifies to a^3 + b^3.
Combine like terms.The mathematical expression a^3 + b^3 is displayed, representing the sum of two cubes.
Sum and Difference of Cubes Pattern

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

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.

a cubed plus b cubed is open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses. a cubed minus b cubed is open parentheses a minus close parentheses open parentheses a squared plus ab plus b squared close parentheses. In both cases, the sign of the first term on the right side of the equation is the same as the sign on the left side of the equation and the sign of the second term is the opposite of the sign on the left side.

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 8.

Table 8
n12345678910
\({n}^{3}\)1827641252163435127291000
Example 9How to Factor the Sum or Difference of Cubes

Factor: \({x}^{3}+64.\)

Check that this is a sum with both terms perfect cubes, then rewrite each term as a cube.

Step 1 is to check if the binomial fits the sum or difference of cubes pattern. For this, we check whether it is a sum or difference. x cubed plus 64 is a sum. Next we check if the first and last terms are perfect cubes. They are Step 2 is to rewrite as cubes. So we rewrite as x cubed plus 4 cubed. Step 3 is to use either the sum or difference of cubes pattern. Since this is a sum of cubes, we get open parentheses x plus 4 close parentheses open parentheses x squared minus 4x plus 4 squared. Step 4 is to simplify inside the parentheses. It is already simplified Step 5 is to check by multiplying the factors.
Try It #17

Factor: \({x}^{3}+27.\)

\((x+3)({x}^{2}-3x+9)\)

Did you get it?
Try It #18

Factor: \({y}^{3}+8.\)

\((y+2)({y}^{2}-2y+4)\)

Did you get it?
Factor the sum or difference of cubes.
  • 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?
  • Write them as cubes.
  • Use either the sum or difference of cubes pattern.
  • Simplify inside the parentheses.
  • Check by multiplying the factors.
Example 10

Factor: \(27{u}^{3}-125{v}^{3}.\)

Rewrite each term as a cube to identify a and b for the difference-of-cubes pattern.

Table 9
The mathematical expression 27u^3 - 125v^3 is displayed in black text on a white background, representing the difference of two cubes.
This binomial is a difference. The first and last
terms are perfect cubes.
Write the terms as cubes.A mathematical image showing the difference of cubes identity: a³ - b³ in red, followed by a specific application of the formula: (3u)³ - (5v)³.
Use the difference of cubes pattern.An image displaying the algebraic identity for the difference of cubes, (a - b)(a^2 + ab + b^2), with a specific example below it where a=3u and b=5v.
Simplify.Two lines of algebraic expressions, illustrating the difference of cubes identity: (a - b)(a^2 + ab + b^2) and (3u - 5v)(9u^2 + 15uv + 25v^2).
Check by multiplying.We’ll leave the check to you.
Try It #19

Factor: \(8{x}^{3}-27{y}^{3}.\)

\((2x-3y)(4{x}^{2}+6xy+9{y}^{2})\)

Did you get it?
Try It #20

Factor: \(1000{m}^{3}-125{n}^{3}.\)

\(125(4{m}^{2} + 2mn + {n}^{2})(2m - n)\)

Did you get it?

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

Example 11

Factor: \(6{x}^{3}y+48{y}^{4}.\)

Factor out the common factor, \(6y,\) before applying the sum-of-cubes pattern.

Table 10
A mathematical expression 6x^3y + 48y^4 is displayed in black text on a white background, featuring variables x and y raised to powers, coefficients, and an addition operator.
Factor the common factor.A mathematical expression is displayed, which reads as 6y multiplied by the sum of x cubed and 8y cubed, written as 6y(x^3 + 8y^3).
This binomial is a sum The first and last
terms are perfect cubes.
Write the terms as cubes.A mathematical expression 6y(x^3 + (2y)^3), with a hint above in red indicating the sum of cubes formula, a^3 + b^3.
Use the sum of cubes pattern.A mathematical expression featuring 6y multiplied by two parenthetical terms: (x + 2y) and (x^2 - x * 2y + (2y)^2
Simplify.The algebraic expression 6y(x + 2y)(x^2 - 2xy + 4y^2), which is a factored form related to the sum of cubes, simplifying to 6y(x^3 + 8y^3).

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.

Try It #21

Factor: \(500{p}^{3}+4{q}^{3}.\)

\(4(5p+q)(25{p}^{2}-5pq+{q}^{2})\)

Did you get it?
Try It #22

Factor: \(432{c}^{3}+686{d}^{3}.\)

\(2(6c+7d)(36{c}^{2}-42cd+49{d}^{2})\)

Did you get it?

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

Example 12

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

Rewrite both terms as cubes: \({(x+5)}^{3}\) and \({(4x)}^{3}.\)

Table 11
The image displays the mathematical expression (x+5)² - 64x³.
This binomial is a difference. The first and
last terms are perfect cubes.
Write the terms as cubes.The image displays the difference of cubes formula a^3 - b^3 in red, above an example expression (x + 5)^3 - (4x)^3, demonstrating its algebraic application.
Use the difference of cubes pattern.A mathematical expression illustrating the difference of cubes factorization (a-b)(a^2+ab+b^2), with 'a' representing (x+5) and 'b' representing 4x.
Simplify.A mathematical expression featuring the product of two polynomials: (x + 5 - 4x) multiplied by (x^2 + 10x + 25 + 4x^2 + 20x + 16x^2).
A mathematical expression showing the product of two polynomials: (-3x + 5) and (21x^2 + 30x + 25).
Check by multiplying.We’ll leave the check to you.
Try It #23

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

\((-2y+1)(13{y}^{2}+5y+1)\)

Did you get it?
Try It #24

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

\((-4n+3)(31{n}^{2}+21n+9)\)

Did you get it?
Media

Access this online resource for additional instruction and practice with factoring special products.

Key Concepts

Section Exercises

Practice Makes Perfect

Factor Perfect Square Trinomials

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

4

\(16{y}^{2}+24y+9\)

\({(4y+3)}^{2}\)

5

\(25{v}^{2}+20v+4\)

6

\(36{s}^{2}+84s+49\)

\({(6s+7)}^{2}\)

7

\(49{s}^{2}+154s+121\)

8

\(100{x}^{2}-20x+1\)

\({(10x-1)}^{2}\)

9

\(64{z}^{2}-16z+1\)

10

\(25{n}^{2}-120n+144\)

\({(5n-12)}^{2}\)

11

\(4{p}^{2}-52p+169\)

12

\(49{x}^{2}+28xy+4{y}^{2}\)

\({(7x+2y)}^{2}\)

13

\(25{r}^{2}+60rs+36{s}^{2}\)

14

\(100{y}^{2}-20y+1\)

\((10y-1{)}^{2}\)

15

\(64{m}^{2}-16m+1\)

16

\(10j{k}^{2}+80jk+160j\)

\(10j{(k+4)}^{2}\)

17

\(64{x}^{2}y-96xy+36y\)

18

\(75{u}^{4}-30{u}^{3}v+3{u}^{2}{v}^{2}\)

\(3{u}^{2}{(5u-v)}^{2}\)

19

\(90{p}^{4}+300{p}^{3}q+250{p}^{2}{q}^{2}\)

Factor Differences of Squares

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

20

\(25{v}^{2}-1\)

\((5v-1)(5v+1)\)

21

\(169{q}^{2}-1\)

22

\(4-49{x}^{2}\)

\((2-7x)(2+7x)\)

23

\(121-25{s}^{2}\)

24

\(6{p}^{2}{q}^{2}-54{p}^{2}\)

\(6{p}^{2}(q-3)(q+3)\)

25

\(98{r}^{3}-72r\)

26

\(24{p}^{2}+54\)

\(6(4{p}^{2}+9)\)

27

\(20{b}^{2}+140\)

28

\(121{x}^{2}-144{y}^{2}\)

\((11x-12y)(11x+12y)\)

29

\(49{x}^{2}-81{y}^{2}\)

30

\(169{c}^{2}-36{d}^{2}\)

\((13c-6d)(13c+6d)\)

31

\(36{p}^{2}-49{q}^{2}\)

32

\(16{z}^{4}-1\)

\((2z-1)(2z+1)(4{z}^{2}+1)\)

33

\({m}^{4}-{n}^{4}\)

34

\(162{a}^{4}{b}^{2}-32{b}^{2}\)

\(2{b}^{2}(3a-2)(3a+2)(9{a}^{2}+4)\)

35

\(48{m}^{4}{n}^{2}-243{n}^{2}\)

36

\({x}^{2}-16x+64-{y}^{2}\)

\((x-8-y)(x-8+y)\)

37

\({p}^{2}+14p+49-{q}^{2}\)

38

\({a}^{2}+6a+9-9{b}^{2}\)

\((a+3-3b)(a+3+3b)\)

39

\({m}^{2}-6m+9-16{n}^{2}\)

Factor Sums and Differences of Cubes

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

40

\({x}^{3}+125\)

\((x+5)({x}^{2}-5x+25)\)

41

\({n}^{6}+512\)

42

\({z}^{6}-27\)

\(({z}^{2}-3)({z}^{4}+3{z}^{2}+9)\)

43

\({v}^{3}-216\)

44

\(8-343{t}^{3}\)

\((2-7t)(4+14t+49{t}^{2})\)

45

\(125-27{w}^{3}\)

46

\(8{y}^{3}-125{z}^{3}\)

\((2y-5z)(4{y}^{2}+10yz+25{z}^{2})\)

47

\(27{x}^{3}-64{y}^{3}\)

48

\(216{a}^{3}+125{b}^{3}\)

\((6a+5b)(36{a}^{2}-30ab+25{b}^{2})\)

49

\(27{y}^{3}+8{z}^{3}\)

50

\(7{k}^{3}+56\)

\(7(k+2)({k}^{2}-2k+4)\)

51

\(6{x}^{3}-48{y}^{3}\)

52

\(2{x}^{2}-16{x}^{2}{y}^{3}\)

\(2{x}^{2}(1-2y)(1+2y+4{y}^{2})\)

53

\(-2{x}^{3}{y}^{2}-16{y}^{5}\)

54

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

\(9(x+1)({x}^{2}+3)\)

55

\({(x+4)}^{3}-27{x}^{3}\)

56

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

\(\text{-}(3y+5)(21{y}^{2}-30y+25)\)

57

\({(y-5)}^{3}+125{y}^{3}\)

Mixed Practice

In the following exercises, factor completely.

58

\(64{a}^{2}-25\)

\((8a-5)(8a+5)\)

59

\(121{x}^{2}-144\)

60

\(27{q}^{2}-3\)

\(3(3q-1)(3q+1)\)

61

\(4{p}^{2}-100\)

62

\(16{x}^{2}-72x+81\)

\({(4x-9)}^{2}\)

63

\(36{y}^{2}+12y+1\)

64

\(8{p}^{2}+2\)

\(2(4{p}^{2}+1)\)

65

\(81{x}^{2}+169\)

66

\(125-8{y}^{3}\)

\((5-2y)(25+10y+4{y}^{2})\)

67

\(27{u}^{3}+1000\)

68

\(45{n}^{2}+60n+20\)

\(5{(3n+2)}^{2}\)

69

\(48{q}^{3}-24{q}^{2}+3q\)

70

\({x}^{2}-10x+25-{y}^{2}\)

\((x-5-y)(x-5+y)\)

71

\({x}^{2}+12x+36-{y}^{2}\)

72

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

\((3x+1)(3{x}^{2}+1)\)

73

\({(y-3)}^{3}-64{y}^{3}\)

Writing Exercises

74

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

Answers will vary.

75

How do you recognize the binomial squares pattern?

76

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

Answers will vary.

77

Maribel factored \({y}^{2}-30y+81\) as \({(y-9)}^{2}.\) Was she right or wrong? How do you know?

Self Check

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

This table has 4 columns 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: factor perfect square trinomials, factor differences of squares, factor sums and differences of cubes. The remaining columns are blank.

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