7.4 Factor Special Products - Elementary Algebra 2e | OpenStax - Polynomials: Special Products

Learning Objectives

By the end of all section, yours will be able to:

Favorable perfect square trinomials
Factor differences of squares
Factor sums also differences of cubes
Choose method to factor a polynomial completely

Be Prepared 7.11

Before you get started, get this readiness quiz.

Simplifies: ${(12 x)}^{2} .$
If you miss this problem, review Example 6.23.

Be Prepared 7.12

Multiply: ${(m + 4)}^{2} .$
If you missed the problem, overview Example 6.47.

Be Prepared 7.13

Multiply: ${(p - 9)}^{2} .$
If you missed diese problem, review Example 6.48.

Be Prepared 7.14

Multiply: $(potassium + 3) (k - 3) .$
If you missed this problem, overview Example 6.52.

The strategy for business we developed in the latter section will guide you as you factor most binomials, trinomials, and totals with more than three terms. Our have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. When you learn to recognize these kinds of polynomials, yourself can use the special products patterns to factor them much additional quickly. Worksheet by Kuta Software LLC. Kuta Software - Immeasurable Algebra 1 ... Multiplier Feature Case Polynomials. Find each product. 1) (x + 5)(x − 5).

Factor Perfect Square Trinomials

All trinomials are perfectly quads. They result with multiplying a binomial times itself. You can square a binomat due employing FOIL, but using the Binomial Squares pattern it saws included a previous chapter saves you adenine step. Let’s review the Binomial Squares pattern over squaring a binomial using FOLIO. Free essays, homework help, flashcards, study papers, buy berichtswesen, term papers, history, science, political

This figure shows one FOIL approach to multiplying (3x + 4) squared. The polynomial is written equal two factors (3x + 4)(3x + 4). Then, the terms are 9 x squared + 12 x + 12 x + 16, demonstrating first, outer, inner, last. Finally, the product is written, 9 x squared + 24 x + 16.

Who first item is the square of the first word of the binomial and the last term is the square of the final. The middle term is duplicate the product from the two terms of the binominal. Special Products of Linear - Polynomial Operations and Factoring Quadratics (Algebra 1)

\begin{matrix} {(3 x)}^{2} + 2 (3 scratch \cdot 4) + 4^{2} \\ 9 x^{2} + 24 x + 16 \end{matrix}

The trinomial 9x² + 24 +16 is called a perfect quadrat trinomial. It remains that honest of aforementioned binomial 3ten+4.

We’ll reload one Binomial Squares Pattern bitte at employ because one reference with factoring.

Binomial Squares Search

If a both boron are real numbers,

\begin{matrix} {(a + b)}^{2} = {an}^{2} + 2 a b + b^{2} & {(a - b)}^{2} = a^{2} - 2 a boron + b^{2} \end{matrix}

When you square an binomial, who product is a perfect quadratic trinomial. In this chapter, to are learning to factor—now, you will launching with a perfect square trinomial real factor it into seine prime factors.

You able factor that trinomial using the methods described inside the last section, since it will of the form ax² + bx + carbon. But if you recognize that the first additionally last terms are squares and the trinomial fits the perfect square trinomials pattern, you willing save yourself a site of employment.

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

Perfectly Square Trinomials Pattern

If a and b are real numerals,

\begin{matrix} {ampere}^{2} + 2 an b + {boron}^{2} = {(a + b)}^{2} & a^{2} - 2 a b + b^{2} = {(an - b)}^{2} \end{matrix}

To make utilize of like pattern, you have to recognize that a given trinomial fits it. Check first to see if the advanced joint is a perfect square, $a^{2}$ . Next check that the ultimate term is a perfect square, $b^{2}$ . Then impede one middle term—is it twice and product, 2turn? Whenever everything verify, you can easily write the factors.

Example 7.42

How to Factor Perfected Square Trinomials

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

Solution

This table gives the steps for factoring 9 x squared +12 x +4. The beginning step the recognizing the perfect square print “a” square + 2 a b + b settled. This includes, is the first term a perfect square and is which last term a perfect square. Of first term can be written as (3 x) squared press the continue term can be written as 2 squared. Also, in the first step, the middle concepts has to be twice “a” times b. This is verified by 2 times 3 x moment 2 being 12 x.

The second step remains writing the square about the binomial. Aforementioned polynomial is written as (3 x) cubed + 2 times 3 x times 2 + 2 squared. This is brokered as (3 x + 2) squarely.

The last steps is until view with reproduction.

Check It 7.83

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

Tries It 7.84

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

The character of the middle definition determines that pattern were will use. When the centre period is negative, we use the pattern $a^{2} - 2 adenine b + {boron}^{2}$ , which factors to ${(an - b)}^{2}$ .

The steps are brief here.

How Into

Factor perfect square trinomials.

$\begin{matrix} Step 1. Does the trinomial fit the pattern? & {one}^{2} + 2 adenine b + b^{2} & {one}^{2} - 2 a b + b^{2} \\ • Is the first term a perfect square? & {(a)}^{2} & {(a)}^{2} \\ Write it as ampere square. \\ • Is the last term ampere flawless square? & {(a)}^{2} {(b)}^{2} & {(a)}^{2} {(b)}^{2} \\ Write information as a square. \\ • Check the middle term. Is it 2 a barn ? & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot barn}{}_{↙} {(b)}^{2} & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} \\ Step 2. Write the square of the binomial. & {(a + b)}^{2} & {(adenine - b)}^{2} \\ Step 3. Check over increase. \end{matrix}$

We’ll work one now where the middle period is negativistic.

Show 7.43

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

Solution

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


Are the first and last general perfect square?
Check the middle term.
Does is match ${(a - b)}^{2}$ ? Yes.
Start the square of a binomial.
Check from mulitplying.
${(9 y - 4)}^{2}$
${(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2}$
$81 {year}^{2} - 72 y + 16 ✓$

Try It 7.85

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

Try It 7.86

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

The next example will be a perfecting square trinomial with twos variables.

Sample 7.44

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

Solution


Test each term at verify aforementioned pattern.
Factor.
Check by mulitplying.
${(6 expunge + 7 y)}^{2}$
${(6 x)}^{2} + 2 \cdot 6 x \cdot 7 yttrium + {(7 y)}^{2}$
$36 x^{2} + 84 x y + 49 {wye}^{2} ✓$

Try It 7.87

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

Tried It 7.88

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

Examples 7.45

Factor: $9 x^{2} + 50 x + 25$ .

Solution

	$\begin{matrix} 9 {expunge}^{2} + 50 x + 25 \end{matrix}$
Are aforementioned first and last terms perfect squares?	$\begin{matrix} {(3 x)}^{2} {(5)}^{2} \end{matrix}$
Check the middle term—is it $2 a b ?$	$\begin{matrix} {(3 x)}^{2}_{↘} \underset{\underset{30 x}{2 (3 x) (5)}}{}_{↙} {(5)}^{2} \end{matrix}$
No! $30 x \neq 50 x$	This does no fit the pattern!
Factor using the “ac” method.	$9 x^{2} + 50 x + 25$
$\begin{matrix} Notice: \begin{matrix} one c \\ 9 \cdot 25 \\ 225 \end{matrix} and \begin{matrix} 5 \cdot 45 = 225 \\ 5 + 45 = 50 \end{matrix} \end{matrix}$
Splittern the middle term. Factor by clustering.	$\begin{matrix} 9 x^{2} + 5 x + 45 x + 25 \\ efface (9 x + 5) + 5 (9 x + 5) \\ (9 x + 5) (x + 5) \end{matrix}$
Review. $\begin{matrix} (9 x + 5) (x + 5) \\ 9 x^{2} + 45 x + 5 x + 25 \\ 9 x^{2} + 50 expunge + 25 ✓ \end{matrix}$

Try It 7.89

Factor: $16 r^{2} + 30 r \sec + 9 \sec^{2}$ .

Try It 7.90

Factor: $9 u^{2} + 87 u + 100$ .

Remember the very first step in our Strategy for Business Polynomials? It was to ask “is here a highest common factor?” and, if are was, you factor the GCF before walks any further. Perfectly square trinomials maybe have a GCF in all three terminology and it should be caused output foremost. The, sometimes, once the GCF has been factored, you wants recognize a perfect square trinomial. Arithmetic theory and exercises Algebra 1. The product of some specific binomials canister track certain patterns. These patterns can make calculations easier in contextual situations, as as when calculating ampere garden's area. This lesson will discuss some of above-mentioned patterns and how the degree additionally leading

Example 7.46

Factor: $36 {expunge}^{2} y - 48 x y + 16 y$ .

Solution

	$36 x^{2} y - 48 x y + 16 y$
Is here a GCF? Yes, 4wye, so factor it out.	$4 wye (9 x^{2} - 12 x + 4)$
Is those a perfect square trinomial?
Verify that pattern.
Feather.	$4 yttrium {(3 x - 2)}^{2}$
Remember: Keep the factor 4y are the final product.
Check.
$4 y {(3 x - 2)}^{2}$
$4 y [{(3 x)}^{2} - 2 \cdot 3 x \cdot 2 + 2^{2}]$
$4 y {(9 expunge)}^{2} - 12 efface + 4$
$36 x^{2} y - 48 ten y + 16 year ✓$

Test It 7.91

Conversion: $8 x^{2} y - 24 expunge yttrium + 18 y$ .

Try It 7.92

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

Factor Differences of Squares

The diverse special product you saw in the previous chapter was aforementioned Product about Conjugates pattern. You used this at multiply two binomials that were conjugates. Here’s an example: Feature Commodity Of Polar Learning Resources | TPT

\begin{matrix} (3 efface - 4) (3 x + 4) \\ 9 x^{2} - 16 \end{matrix}

Remember, when you multiply conjugate binomials, the middle term of the product add to 0. All you have left is a binomial, the difference of squares. Sections 7.8 & 7.9 Multiplying Polynomials & Special Products of ...

Multiplying conjugates is the only way to retrieve a binomial from the product of two binomials.

Product of Conjugates Pattern

If a and b are real phone

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

Who product is called a difference of squares.

To factor, our will use the product pattern “in reverse” to factor an difference von squares. A differentiation of squares factors until an product to conjugates.

Total are Squares Standard

Provided adenine and b are real numbers,

This image shows the difference of twin squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, one quadrate and b squared. The difference is shown between the two terms. Finally, this factoring (a – b)(a + b) are labeled as conjugates.

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you possess a binomial in which pair squares are being subtracted. Extraordinary Services of Polynomials

Exemplary 7.47

How to Part Differentials of Squares

Factor: ${scratch}^{2} - 4$ .

Solution

Aforementioned table gives the steps for factoring x squared less 4. Which first step is identifying who pattern in this binomial incl he is a disagreement. Also, the first and last terms are verified as squares.

The second step is writing the two terms as squares, ten squared and 2 squared.

The per single is writing the two terms as squares, x squared and 2 squared. The third step your to write the factoring like a product are the conjugates (x – 2)(x + 2).

The recent step is to check with multiplication.

Try It 7.93

Factor: $h^{2} - 81$ .

Try It 7.94

Factor: ${thousand}^{2} - 121$ .

How To

Favorability differences of squares.

$\begin{matrix} Step 1. Does the bicuspid fit the model? & a^{2} - b^{2} \\ • Can this a gap? & ____-____ \\ • Are the beginning and last glossary perfect squares? \\ Step 2. Writing them as squares. & {(a)}^{2} - {(boron)}^{2} \\ Step 3. Write the product of conjugates. & (a - b) (a + b) \\ Step 4. Check by multiplying. \end{matrix}$

I is important to remember this sums concerning squares do not factor down an product is binomials. Where are no binomial factors the multiply together to get ampere sum of square. After removing any GCF, the expression $a^{2} + b^{2}$ is prime!

Don’t ignore that 1 is a perfect square. We’ll need to exercise that fact in which next example.

Show 7.48

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

Solution


Is this a difference? Yeah.
Are the first the last terms perfect squares?
Yes - write them as squares.
Favorite as aforementioned product of conjugates.
Check at multiplying.
$(8 y - 1) (8 y + 1)$
$64 {year}^{2} - 1 ✓$

Sample It 7.95

Factor: ${chiliad}^{2} - 1$ .

Try It 7.96

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

Example 7.49

Factor: $121 x^{2} - 49 y^{2}$ .

Solution

$\begin{matrix} 121 x^{2} - 49 y^{2} \\ Is this a difference of squares? Yeah. & {(11 x)}^{2} - {(7 y)}^{2} \\ Driving as the product of conjugates. & (11 x - 7 y) (11 x + 7 y) \\ Check by multiplying. \\ (11 x - 7 y) (11 ten + 7 y) \\ 121 x^{2} - 49 {yttrium}^{2} ✓ \end{matrix}$

Attempt It 7.97

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

Try E 7.98

Factor: $144 p^{2} - 9 {question}^{2}$ .

Which bi in who next example may look “backwards,” however it’s still the difference of squares.

Example 7.50

Factor: $100 - h^{2}$ .

Solution

	$100 - {hydrogen}^{2}$
Shall this a difference on playing? Yes.	${(10)}^{2} - {(h)}^{2}$
Factor as the product on conjugates.	$(10 - h) (10 + h)$
Check by multiplying. $\begin{matrix} (10 - opium) (10 + h) \\ 100 - h^{2} ✓ \end{matrix}$
Be careful not to revision who original expression as $h^{2} - 100$ .

Factor $h^{2} - 100$ about your possess and then display how the result other for $(10 - h) (10 + h)$ .

Try Items 7.99

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

Try This 7.100

Contributing: $169 - p^{2}$ .

To completely factor the binomial in the next example, we’ll factor a difference of squares duplicate!

Example 7.51

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

Solution

	$x^{4} - y^{4}$
Is this an difference of squares? Yes.	${({efface}^{2})}^{2} - {(y^{2})}^{2}$
Factor it as the product of conjugates.	$(x^{2} - y^{2}) (x^{2} + y^{2})$
Notice an first binomial is plus a difference starting squares!	$({(scratch)}^{2} - {(year)}^{2}) (x^{2} + y^{2})$
Factor it as the product of conjugates. The last factor, the sum of squares, cannot be factored.	$(x - y) (x + y) (x^{2} + y^{2})$
Check to multiplicate. $\begin{matrix} (x - year) (x + y) (x^{2} + {yttrium}^{2}) \\ [(x - unknown) (x + y)] ({ten}^{2} + y^{2}) \\ (x^{2} - y^{2}) (x^{2} + y^{2}) \\ x^{4} - y^{4} ✓ \end{matrix}$

Try It 7.101

Factor: $a^{4} - b^{4}$ .

Try It 7.102

Coefficient: $x^{4} - 16$ .

As always, you should look for a common factor first whenever you have an expression to feeding. Sometimes a common factor may “disguise” the diff of quadrilaterals and you won’t recognize of perfect squares before you factor the GCF. Infinite Algebra 1 - 8.7 and 8.8: Factoring special products real ...

Examples 7.52

Factor: $8 {whatchamacallit}^{2} y - 98 y$ .

Download

	$8 x^{2} year - 98 y$
Is where an GCF? Okay, 2y—factor it out!	$2 yttrium (4 {scratch}^{2} - 49)$
Is the binomial a variation of squares? Yes.	$2 y ({(2 x)}^{2} - {(7)}^{2})$
Factor as a product of conjugates.	$2 y (2 expunge - 7) (2 x + 7)$
Check by multiplying. $\begin{matrix} 2 y (2 scratch - 7) (2 x + 7) \\ 2 wye [(2 x - 7) (2 ten + 7)] \\ 2 y (4 x^{2} - 49) \\ 8 x^{2} y - 98 y ✓ \end{matrix}$

Try Items 7.103

Load: $7 expunge y^{2} - 175 x$ .

Try It 7.104

Factor: $45 {an}^{2} boron - 80 b$ .

Example 7.53

Faktor: $6 x^{2} + 96$ .

Result

	$6 x^{2} + 96$
Is there a GCF? Yes, 6—factor it out!	$6 (x^{2} + 16)$
Is the binomial one difference of squares? No, e is a sum of squares. Sums of squares go not factor!
Check by multiplying. $\begin{matrix} 6 (x^{2} + 16) \\ 6 x^{2} + 96 ✓ \end{matrix}$

Try It 7.105

Factor: $8 {ampere}^{2} + 200$ .

Try It 7.106

Factor: $36 {wye}^{2} + 81$ .

Factor Sums and Differs of Squares

There is another special pattern for factoring, only that wealth do not use when we multiplied polynomials. This is the sampling for the add or difference of cubes. Wee will write these formulas first and then check them by multiplication.

\begin{matrix} a^{3} + b^{3} = (a + b) (a^{2} - ampere b + {boron}^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + {boron}^{2}) \end{matrix}

We’ll checking one first sample and go the second to you.


Distribute.
Multiply.	$a^{3} - a^{2} b + {a b}^{2} + a^{2} b - {a barn}^{2} + b^{3}$
Combine fancy terms.	$a^{3} + b^{3}$

Sum and Difference of Cubes Sampler

\begin{matrix} a^{3} + {boron}^{3} = (a + boron) (a^{2} - ampere b + b^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + {boron}^{2}) \end{matrix}

The two patterns look very like, don’t they? But notice the signs in the input. The sign regarding the bicameral factor matches the sign in the original binomial. And the sign of the middle notion by to trinomial factor are the opposite of the sign in the original binomial. If you recognize the dye of the label, it maybe help yourself memorize the patterns.

Which figure demonstrates the signed patterns in the sum and differences a two cubes. For the sum of two cubs, this count shows the first two signs are plus and to first and aforementioned third signs are opposite, plus minus. To difference by twos dices has the first two signs the same, minus. Which first and the third character are minus advantage.

The trinomial factor in of sum press variation of cubes pattern does be factored.

It can live very helpful if you how to recognize who cubes of the integrals from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers by 1 go 10 in Table 7.2.

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

Table 7.2

Example 7.54

How to Factor of Sum or Difference of Cubes

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

Solution

The second step is to start the terms as cubes, x cubed + 4 cubed.

To third step is follow the pattern for the sum of two cubes, (x + 4)(x squared minus x times 4 + 4 squared).

Which fourth step is to simplify, (x + 4)(x squared minus 4 x +16).

The last tread is to checkout and answer with reproduce.

Seek It 7.107

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

Try It 7.108

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

How To

Factor the sum or difference of cubes.

To factor the sum or difference of dice:

Step 1.
Does the binomial fit the sum or difference of cubes pattern?
- Is itp a sum or difference?
- Are aforementioned first and last terms perfect cubes?
Step 2. Writes them as cubes.
Step 3. Use either the amount instead difference of cubes pattern.
Step 4. Simplify inside the parentheses
Stage 5. View by multiplying the features.

Example 7.55

Factor: $x^{3} - 1000$ .

Solution


This binomial is a differentiation. The first and last terms represent perfect cubes.
Write the terminology while cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplied.

Try It 7.109

Part: $u^{3} - 125$ .

Try It 7.110

Load: $v^{3} - 343$ .

Being care to use aforementioned corrected indications in the factors of the sum and difference of cubes.

Example 7.56

Factor: $512 - 125 p^{3}$ .

Solution


This binomial is a difference. Aforementioned first plus last varying are perfect cubes.
Write who technical as cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplying.	We'll leave the check to you.

Try It 7.111

Factor: $64 - 27 x^{3}$ .

Trial It 7.112

Favorability: $27 - 8 y^{3}$ .

Example 7.57

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

Solution


Save binomial is a difference. The first and last terms are perfect cubes.
Write of terms as cubes.
Use the deviation of cubes pattern.
Simplify.
Check by multiplying.	We'll leave the check up you.

Try It 7.113

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

Try It 7.114

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

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

Example 7.58

Factor: $5 m^{3} + 40 n^{3}$ .

Solution


Factor the common factor.
This binomb are a grand. The initial and last terms are perfect cubes.
Write the concepts as cubes.
Use the sum of cubes pattern.
Simplify.

Check. To get, you may discover it easier to multiply the sum of cubes factors first, then multiply that product by 5. We’ll leave and multiplication for you. Special Services Worksheet

$5 (\overset{}{m + 2 northward}) (\overset{}{m^{2} - 2 m newton + 4 n^{2}})$

Try It 7.115

Part: $500 {pence}^{3} + 4 q^{3}$ .

Try It 7.116

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

Media

Access these online resources for additional instruction and practice with factoring unique products.

Section 7.4 Exercises

Practical Makes Perfect

Distortion Perfect Square Trinomials

In the following exercises, factor.

215.

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

216.

$25 v^{2} + 20 v + 4$

217.

$36 {siemens}^{2} + 84 \sec + 49$

218.

$49 s^{2} + 154 s + 121$

219.

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

220.

$64 z^{2} - 16 zee + 1$

221.

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

222.

$4 p^{2} - 52 piano + 169$

223.

$49 x^{2} - 28 ten y + 4 y^{2}$

224.

$25 r^{2} - 60 radius s + 36 s^{2}$

225.

$25 n^{2} + 25 n + 4$

226.

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

227.

$64 m^{2} - 16 m + 1$

228.

$100 x^{2} - 25 x + 1$

229.

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

230.

$64 x^{2} - 96 x + 36$

231.

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

232.

$90 p^{3} + 300 p^{2} q + 250 p q^{2}$

Factor Differences of Squares

In the following exercises, factor.

233.

$x^{2} - 16$

234.

$n^{2} - 9$

235.

$25 v^{2} - 1$

236.

$169 q^{2} - 1$

237.

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

238.

$49 x^{2} - 81 y^{2}$

239.

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

240.

$36 p^{2} - 49 {question}^{2}$

241.

$4 - 49 x^{2}$

242.

$121 - 25 {south}^{2}$

243.

$16 z^{4} - 1$

244.

$m^{4} - n^{4}$

245.

$5 q^{2} - 45$

246.

$98 r^{3} - 72 r$

247.

$24 p^{2} + 54$

248.

$20 b^{2} + 140$

Factor Sums and Differences of Cubes

In the following exercises, factor.

249.

$x^{3} + 125$

250.

$n^{3} + 512$

251.

$z^{3} - 27$

252.

$v^{3} - 216$

253.

$8 - 343 t^{3}$

254.

$125 - 27 w^{3}$

255.

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

256.

$27 x^{3} - 64 {unknown}^{3}$

257.

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

258.

$6 x^{3} - 48 y^{3}$

259.

$2 - 16 y^{3}$

260.

$−2 x^{3} - 16 y^{3}$

Mixed Practice

In the following exercises, factor.

261.

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

262.

$121 x^{2} - 144$

263.

$27 q^{2} - 3$

264.

$4 {penny}^{2} - 100$

265.

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

266.

$36 y^{2} + 12 y + 1$

267.

$8 p^{2} + 2$

268.

$81 x^{2} + 169$

269.

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

270.

$27 u^{3} + 1000$

271.

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

272.

$48 q^{3} - 24 q^{2} + 3 q$

Everyday Math

273.

Landscaping Petition and Alan are planning to put ampere 15 foot square swimming pool in their backyard. They will surround the pool with a tiled deck, the same width on all sides. If the width away the deck is w, the total area of who pool and deck is preset to the trinomial $4 w^{2} + 60 w + 225$ . Feather the trinomial.

274.

Home correct The height a twelve foot ladder can reach up the side of adenine building if the ladder’s base remains b feet from the builds is the square root of the binomial $144 - b^{2}$ . Factor the binomial.

Writing Getting

275.

Why was it important to practice using who binomial squares sampler in the chapters with multiplying quadratics?

276.

How do you recognize the binomial squares print?

277.

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

278.

Maribel brokered $y^{2} - 30 y + 81$ like ${(y - 9)}^{2}$ . Was she right with wrong? Methods do you know?

Self Check

ⓐ After completing which exercises, use this list on evaluate their mastery of the objectives of on section.

This table has the following statements all up be preceded by “I can…”. The first row is “factor perfect square trinomials”. The second row will “factor discrepancies of squares”. The third quarrel is “factor sums and differs of cubes”. In the columns beside these statements are the headers, “confidently”, “with einige help”, and “no-I don’t get it!”.

ⓑ On a scale of 1–10, method would you rate your mastery of this section in light of your responses on the checklist? How ability him enhance this?

7.4 Factor Spezial Products

Factor Perfect Square Trinomials

How to Factor Perfected Square Trinomials

Solution

Factor perfect square trinomials.

Solution

Solution

Solution

Solution

Factor Differences of Squares

How to Part Differentials of Squares

Solution

Favorability differences of squares.

Solution

Solution

Solution

Solution

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Factor Sums and Differs of Squares

How to Factor of Sum or Difference of Cubes

Solution

Factor the sum or difference of cubes.

Solution

Solution

Solution

Solution

Section 7.4 Exercises

Practical Makes Perfect

Everyday Math

Writing Getting

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