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Section P.4 Factoring Polynomials

Before Starting This Section, Review

1 Exponents (Section P.2 )
2 Polynomials (Section P.3 )
3 Value of algebraic expressions (Section P.1 , page 11)

Objectives

1 Identify and factor out the greatest common monomial factor.
2 Factor trinomials with a leading coefficient of 1.
3 Factor perfect-square trinomials.
4 Factor the difference of squares.
5 Factor the difference and sum of cubes.
6 Factor by grouping.
7 Factor trinomials.

Profit from Research and Development

The finance department of a major drug company projects that an investment of x million dollars in research and development for a specific type of vaccine will return a profit (or loss) of (0.012x3−32.928) $(0.012 x^{3} - 32.928)$ million dollars.

The company would have a “startup” cost of $12 million and is prepared to invest up to $24 million at a rate of $1 million per year. You could probably stare at this polynomial for a long time without getting a clear picture of what kind of return to expect from different investments. In Example 7, we use the methods of this section to rewrite this profit–loss polynomial, making it easy to explain the result of investing various amounts.

The Greatest Common Monomial Factor

1 Identify and factor out the greatest common monomial factor.

The process of factoring polynomials that we study in this section “reverses” the process of multiplying polynomials that we studied in the previous section. To factor any sum of terms means to write it as a product of factors. Consider the product

(x + 3) (x - 3) = x 2 - 9 .

$(x + 3) (x - 3) = x^{2} - 9 .$

The polynomials (x+3) $(x + 3)$ and (x−3) $(x - 3)$ are called factors of the polynomial x2−9. $x^{2} - 9 .$ So, factoring is the opposite of multiplying. See the margin. When factoring, we start with one polynomial and write it as a product of other polynomials. The polynomials in the product are the factors of the given polynomial.

The first thing to look for when factoring a polynomial is a factor that is common to every term. This common factor can be “factored out” by the distributive property, ab+ac=a(b+c). $a b + a c = a (b + c) .$

Factoring Out a Monomial

In this section, we are concerned only with polynomials having integer coefficients. This restriction is called factoring over the integers.

Polynomial	Common Factor	Factored Form
9+18y $9 + 18 y$	9	9(1+2y) $9 (1 + 2 y)$
2y2+6y $2 y^{2} + 6 y$	2`y`	2y(y+3) $2 y (y + 3)$
7x4+3x3+x2 $7 x^{4} + 3 x^{3} + x^{2}$	x2 $x^{2}$	x2(7x2+3x+1) $x^{2} (7 x^{2} + 3 x + 1)$
5x+3 $5 x + 3$	No common factor	5x+3 $5 x + 3$

You can verify that any factorization is correct by multiplying the factors. We factored these polynomials by finding the greatest common monomial factor of their terms.

Example 1 Factoring by Using the GCF

Factor.

16x3+24x2 $16 x^{3} + 24 x^{2}$
5x4+20x2+25x $5 x^{4} + 20 x^{2} + 25 x$
x2(x−3)+7(x−3) $x^{2} (x - 3) + 7 (x - 3)$

Solution

The greatest integer that divides both 16 and 24 is 8. The exponents on x are 3 and 2, so 2 is the least exponent. So the GCF of 16x3+24x2 $16 x^{3} + 24 x^{2}$ is 8x2. $8 x^{2} .$ We write

$16 x 3 + 24 x 2 = = 8 x 2 (2 x) + 8 x 2 (3) . 8 x 2 (2 x + 3) Write each term as the product of the GCF and another factor . Use the distributive property to factor out the GCF .$ $\begin{array}{rcll} 16 x^{3} + 24 x^{2} & = & 8 x^{2} (2 x) + 8 x^{2} (3) . & Write each term as the product of \\ the GCF and another factor . \\ = & 8 x^{2} (2 x + 3) & Use the distributive property to \\ factor out the GCF . \end{array}$

We can check the results of factoring by multiplying.

Check: 8x2(2x+3)=8x2(2x)+8x2(3)=16x3+24x2. $8 x^{2} (2 x + 3) = 8 x^{2} (2 x) + 8 x^{2} (3) = 16 x^{3} + 24 x^{2} .$
The greatest integer that divides 5, 20, and 25 is 5. The exponents on x are 4, 2, and 1 (x=x1), $(x = x^{1}),$ so 1 is the least exponent. So the GCF of 5x4+20x2+25x $5 x^{4} + 20 x^{2} + 25 x$ is 5x. We write

$5 x 4 + 20 x 2 + 25 x = 5 x (x 3) + 5 x (4 x) + 5 x (5) = 5 x (x 3 + 4 x + 5) .$ $5 x^{4} + 20 x^{2} + 25 x = 5 x (x^{3}) + 5 x (4 x) + 5 x (5) = 5 x (x^{3} + 4 x + 5) .$

You should check this result by multiplying.
In this case, the GCF is the common binomial factor (x−3). $(x - 3) .$ So,

$x 2 (x - 3) + 7 (x - 3) = (x - 3) (x 2 + 7) Factor out the common binomial factor .$ $\begin{array}{l} x^{2} (x - 3) + 7 (x - 3) = (x - 3) (x^{2} + 7) & Factor out the common \\ binomial factor . \end{array}$

Practice Problem 1

Factor.
1. 6x5+14x3 $6 x^{5} + 14 x^{3}$
2. 7x5+21x4+35x2 $7 x^{5} + 21 x^{4} + 35 x^{2}$
3. 5x2(x−y)+2(x−y) $5 x^{2} (x - y) + 2 (x - y)$

Polynomials that cannot be factored as a product of two polynomials (excluding the constant polynomials 1 and −1 $- 1$ ) are said to be irreducible. A polynomial is said to be factored completely when it is written as a product consisting of only irreducible factors. By agreement, the greatest integer common to all of the terms in the polynomial is not factored. We also allow repeated irreducible factors to be written as a power of a single factor.

Factoring Trinomials of the Form x2+Bx+C $x^{2} + B x + C$

2 Factor trinomials with a leading coefficient of 1.

Recall that (x+a)(x+b)=x2+(a+b)x+ab. $(x + a) (x + b) = x^{2} + (a + b) x + a b .$

We reverse the sides of this equation to get the form used for factoring

x 2 + (a + b) x + a b = (x + a) (x + b) .

$x^{2} + (a + b) x + a b = (x + a) (x + b) .$

Side Note

To factor a trinomial in standard form with a leading coefficient of 1, we need two integers a and b whose product is the constant term and whose sum is the coefficient of the middle term.

Example 2 Factoring x2+Bx+C $x^{2} + B x + C$ by Using the Factors of `C`

Factor.

x2+8x+15 $x^{2} + 8 x + 15$
x2−6x−16 $x^{2} - 6 x - 16$
x2+x+2 $x^{2} + x + 2$

Solution

To factor x2+Bx+C, $x^{2} + B x + C,$ we look for two integers a and b such that ab=C $a b = C$ and a+b=B. $a + b = B .$

We want integers a and b such that ab=15 $a b = 15$ and a+b=8. $a + b = 8 .$

Because the factors 3 and 5 of 15 in the third column have a sum of 8,

$x 2 + 8 x + 15 = (x + 3) (x + 5) .$ $x^{2} + 8 x + 15 = (x + 3) (x + 5) .$

You should check this answer by multiplying.
We must find two integers a and b with ab=−16 $a b = - 16$ and a+b=−6. $a + b = - 6 .$

Because the factors 2 and −8 $- 8$ of −16 $- 16$ in the fifth column give a sum of −6, $- 6,$

$x 2 - 6 x - 16 = (x + 2) [x + (- 8)] = (x + 2) (x - 8) .$ $x^{2} - 6 x - 16 = (x + 2) [x + (- 8)] = (x + 2) (x - 8) .$

You should check this answer by multiplying.

We must find two integers a and b with ab=2 $a b = 2$ and a+b=1. $a + b = 1 .$

Factors of 2	1, 2	−1,−2 $- 1, - 2$
Sum of factors	3	−3 $- 3$

The coefficient of the middle term, 1, does not appear as the sum of any of the factors of the constant term, 2. Therefore, x2+x+2 $x^{2} + x + 2$ is irreducible.

Practice Problem 2

Factor.
1. x2+6x+8 $x^{2} + 6 x + 8$
2. x2−3x−10 $x^{2} - 3 x - 10$

Factoring Formulas

We investigate how other types of polynomials can be factored by reversing the product formulas on page 37.

Perfect-Square Trinomials

3 Factor perfect-square trinomials.

To factor a perfect square Trinomial, we use the special products:

A 2 + 2 A B + B 2 = (A + B) 2 and A 2 - 2 A B + B 2 = (A - B) 2

$A^{2} + 2 A B + B^{2} = {(A + B)}^{2} and A^{2} - 2 A B + B^{2} = {(A - B)}^{2}$

Example 3 Factoring a Perfect-Square Trinomial

Factor.

x2+10x+25 $x^{2} + 10 x + 25$
16x2−8x+1 $16 x^{2} - 8 x + 1$

Solution

The first term, x2, $x^{2},$ and the third term, 25=52, $25 = 5^{2},$ are perfect squares. Further, the middle term is twice the product of the terms being squared, that is, 10x=2(5x). $10 x = 2 (5 x) .$

$x 2 + 10 x + 25 = x 2 + 2 \cdot 5 \cdot x + 52 = (x + 5) 2$ $x^{2} + 10 x + 25 = x^{2} + 2 \cdot 5 \cdot x + 5^{2} = {(x + 5)}^{2}$
The first term, 16x2=(4x)2, $16 x^{2} = {(4 x)}^{2},$ and the third term, 1=12, $1 = 1^{2},$ are perfect squares. Further, the middle term is the negative of twice the product of terms being squared, that is, −8x=−2(4x)(1). $- 8 x = - 2 (4 x) (1) .$

$16 x 2 - 8 x + 1 = (4 x) 2 - 2 (4 x) (1) + 12 = (4 x - 1) 2$ $16 x^{2} - 8 x + 1 = {(4 x)}^{2} - 2 (4 x) (1) + 1^{2} = {(4 x - 1)}^{2}$

Practice Problem 3

Factor.
1. x2+4x+4 $x^{2} + 4 x + 4$
2. 9x2−6x+1 $9 x^{2} - 6 x + 1$

Side Note

You should check your answer to any factoring problem by multiplying the factors to verify that the result is the original expression.

Difference of Squares

4 Factor the difference of squares.

To factor the difference of squares, we use the reverse of the special product A2−B2=(A+B)(A−B). $A^{2} - B^{2} = (A + B) (A - B) .$

Example 4 Factoring the Difference of Squares

Factor.

x2−4 $x^{2} - 4$
25x2−49 $25 x^{2} - 49$

Solution

We write x2−4 $x^{2} - 4$ as the difference of squares.

$x 2 - 4 = x 2 - 22 = (x + 2) (x - 2)$ $x^{2} - 4 = x^{2} - 2^{2} = (x + 2) (x - 2)$
We write 25x2−49 $25 x^{2} - 49$ as the difference of squares.

$25 x 2 - 49 = (5 x) 2 - 72 = (5 x + 7) (5 x - 7)$ $25 x^{2} - 49 = {(5 x)}^{2} - 7^{2} = (5 x + 7) (5 x - 7)$

Practice Problem 4

Factor.
1. x2−16 $x^{2} - 16$
2. 4x2−25 $4 x^{2} - 25$

What about the sum of two squares? Let’s try to factor x2+22=x2+4. $x^{2} + 2^{2} = x^{2} + 4 .$ Notice that x2+4=x2+0⋅x+4 $x^{2} + 4 = x^{2} + 0 \cdot x + 4$ so that the middle term is 0=0⋅x. $0 = 0 \cdot x .$ We need two factors, a and b, of 4 whose sum is zero (a+b=0). $(a + b = 0) .$

Factors of 4	1, 4	−1,−4 $- 1, - 4$	2, 2	−2,−2 $- 2, - 2$
Sum of factors	5	−5 $- 5$	4	−4 $- 4$

Because no sum of factors is 0, x2+4 $x^{2} + 4$ is irreducible. A similar result holds regardless of what integer replaces 2 in the process. Consequently, x2+a2 $x^{2} + a^{2}$ is irreducible for any integer a.

Example 5 Using the Difference of Squares Twice

Factor x4−16. $x^{4} - 16 .$

Solution

We write x4−16 $x^{4} - 16$ as the difference of squares.

x 4 - 16 = (x 2) 2 - 42 = (x 2 + 4) (x 2 - 4)

$x^{4} - 16 = {(x^{2})}^{2} - 4^{2} = (x^{2} + 4) (x^{2} - 4)$

From Example 4, part a, we know that x2−4=(x+2)(x−2). $x^{2} - 4 = (x + 2) (x - 2) .$ Consequently,

x 4 - 16 = = (x 2 + 4) (x 2 - 4) (x 2 + 4) (x + 2) (x - 2) Replace x 2 - 4 with (x + 2) (x - 2) .

$\begin{array}{rcll} x^{4} - 16 & = & (x^{2} + 4) (x^{2} - 4) \\ = & (x^{2} + 4) (x + 2) (x - 2) & Replace x^{2} - 4 with (x + 2) (x - 2) . \end{array}$

Because x2+4,x+2, $x^{2} + 4, x + 2,$ and x−2 $x - 2$ are irreducible, x4−16 $x^{4} - 16$ is completely factored.

Practice Problem 5

Factor x4−81. $x^{4} - 81 .$

Difference and Sum of Cubes

5 Factor the difference and sum of cubes.

We can also use the reverse of the special-product formulas to factor the difference and sum of cubes.

Example 6 Factoring the Difference and Sum of Cubes

Factor.

x3−64 $x^{3} - 64$
8x3+125 $8 x^{3} + 125$

Solution

We write x3−64 $x^{3} - 64$ as the difference of cubes.

$x 3 - 64 = x 3 - 43 = (x - 4) (x 2 + 4 x + 16)$ $x^{3} - 64 = x^{3} - 4^{3} = (x - 4) (x^{2} + 4 x + 16)$
We write 8x3+125 $8 x^{3} + 125$ as the sum of cubes.

$8 x 3 + 125 = (2 x) 3 + 53 = (2 x + 5) (4 x 2 - 10 x + 25)$ $8 x^{3} + 125 = {(2 x)}^{3} + 5^{3} = (2 x + 5) (4 x^{2} - 10 x + 25)$

Practice Problem 6

Factor.
1. x3−125 $x^{3} - 125$
2. 27x3+8 $27 x^{3} + 8$

Example 7 Determining Profit from Research and Development

In the beginning of this section, we introduced the polynomial 0.012x3−32.928 $0.012 x^{3} - 32.928$ that gave the profit (or loss) a drug company would have after an investment of x million dollars. If the company had an initial investment of $12 million and invests an additional $1 million each year, determine when this investment returns a profit. What will the profit (or loss) be in six years?

Solution

First, notice that 0.012 is a common factor of both terms of the polynomial because 32.928=(0.012)(2744). $32.928 = (0.012) (2744) .$ We now factor 0.012x3−32.928 $0.012 x^{3} - 32.928$ as follows:

0.012 x 3 - 32.928 = = = = 0.012 x 3 - (0.012) (2744) 0.012 (x 3 - 2744) 0.012 (x 3 - 143) 0.012 (x - 14) (x 2 + 14 x + 196) Factor out 0.012. 2744 = 143 Difference of cubes

$\begin{array}{rcll} 0.012 x^{3} - 32.928 & = & 0.012 x^{3} - (0.012) (2744) \\ = & 0.012 (x^{3} - 2744) & Factor out 0.012. \\ = & 0.012 (x^{3} - 14^{3}) & 2744 = 14^{3} \\ = & 0.012 (x - 14) (x^{2} + 14 x + 196) & Difference of cubes \end{array}$

This product has three factors, and the factor x2+14x+196 $x^{2} + 14 x + 196$ is positive for positive values of x because each term is positive. The sign of the factor x−14 $x - 14$ then determines the sign of the entire polynomial. The factor x−14 $x - 14$ is negative if x is less than 14, zero if x equals 14, and positive if x is greater than 14. Consequently, the company receives a profit when x is greater than 14. Because the initial investment is $12 million, with an additional $1 million invested each year, it will be two years before the company breaks even on this venture. Every year thereafter it will profit from the investment. In six years, the company will have invested the initial $12 million plus an additional $6 million ($1 million per year). So the total investment is $18 million. To find the profit (or loss) with $18 million invested, we let x=18 $x = 18$ in the profit–loss polynomial: 0.012(18)3−32.928=$37.056 million $0.012 {(18)}^{3} - 32.928 = $ 37.056 million$ .

The company will have made a profit of over $37 million in six years.

Practice Problem 7

In Example 7 , what will the profit be in four years?

Factoring by Grouping

6 Factor by grouping.

For polynomials having four terms and a GCF of 1, when none of the preceding factoring techniques apply, we can sometimes group the terms in such a way that each group has a common factor. This technique is called factoring by grouping.

Example 8 Factoring by Grouping

Factor.

x3+2x2+3x+6 $x^{3} + 2 x^{2} + 3 x + 6$
6x3−3x2−4x+2 $6 x^{3} - 3 x^{2} - 4 x + 2$
x2+4x+4−y2 $x^{2} + 4 x + 4 - y^{2}$

Solution

x3+2x2+3x+6===(x3+2x2)+(3x+6)x2(x+2)+3(x+2)(x+2)(x2+3)Group terms so that eachgroup can be factored.Factor the grouped binomials.Factor out the commonbinomial factor x+2. $\begin{array}{rcll} x^{3} + 2 x^{2} + 3 x + 6 & = & (x^{3} + 2 x^{2}) + (3 x + 6) & Group terms so that each \\ group can be factored . \\ = & x^{2} (x + 2) + 3 (x + 2) & Factor the grouped binomials . \\ = & (x + 2) (x^{2} + 3) & Factor out the common \\ binomial factor x + 2. \end{array}$
6x3−3x2−4x+2===(6x3−3x2)+(−4x+2)3x2(2x−1)+(−2)(2x−1)(2x−1)(3x2−2)Group terms so that eachgroup can be factored.Factor the groupedbinomials.Factor out the commonbinomial factor 2x−1. $\begin{array}{rcll} 6 x^{3} - 3 x^{2} - 4 x + 2 & = & (6 x^{3} - 3 x^{2}) + (- 4 x + 2) & Group terms so that each \\ group can be factored . \\ = & 3 x^{2} (2 x - 1) + (- 2) (2 x - 1) & Factor the grouped \\ binomials . \\ = & (2 x - 1) (3 x^{2} - 2) & Factor out the common \\ binomial factor 2 x - 1. \end{array}$
x2+4x+4−y2===(x2+4x+4)−y2(x+2)2−y2(x+2−y)(x+2+y)Group terms so that each groupcan be factored.Factor the trinomial.Difference of squares $\begin{array}{rcll} x^{2} + 4 x + 4 - y^{2} & = & (x^{2} + 4 x + 4) - y^{2} & Group terms so that each group \\ can be factored . \\ = & {(x + 2)}^{2} - y^{2} & Factor the trinomial . \\ = & (x + 2 - y) (x + 2 + y) & Difference of squares \end{array}$

Practice Problem 8

Factor.
1. x3+3x2+x+3 $x^{3} + 3 x^{2} + x + 3$
2. 28x3−20x2−7x+5 $28 x^{3} - 20 x^{2} - 7 x + 5$
3. x2−y2−2y−1 $x^{2} - y^{2} - 2 y - 1$

Example 9 Factoring by Making a Perfect Square

Factor: x4+3x2+4 $x^{4} + 3 x^{2} + 4$

Solution

Notice that a change in the middle term would give the perfect square (x2+2)2=x4+4x2+4. ${(x^{2} + 2)}^{2} = x^{4} + 4 x^{2} + 4 .$

x 4 + 3 x 2 + 4 = x 4 + 4 x 2 - x 2 + 4 = x 4 + 4 x 2 + 4 - x 2 = (x 2 + 2) 2 - x 2 = (x 2 + 2 + x) (x 2 + 2 - x) = (x 2 + x + 2) (x 2 - x + 2) 3 x 2 = 4 x 2 - x 2 Rewrite . x 4 + 4 x 2 + 4 = (x 2 + 2) 2 Difference of two squares Rewrite in standard form .

$\begin{array}{l} x^{4} + 3 x^{2} + 4 \\ = x^{4} + 4 x^{2} - x^{2} + 4 & 3 x^{2} = 4 x^{2} - x^{2} \\ = x^{4} + 4 x^{2} + 4 - x^{2} & Rewrite . \\ = {(x^{2} + 2)}^{2} - x^{2} & x^{4} + 4 x^{2} + 4 = {(x^{2} + 2)}^{2} \\ = (x^{2} + 2 + x) (x^{2} + 2 - x) & Difference of two squares \\ = (x^{2} + x + 2) (x^{2} - x + 2) & Rewrite in standard form . \end{array}$

Practice Problem 9

Factor: x4+5x2+9 $x^{4} + 5 x^{2} + 9$

Factoring Trinomials of the Form Ax2+Bx+C $A x^{2} + B x + C$

7 Factor trinomials.

To factor the trinomial Ax2+Bx+C $A x^{2} + B x + C$ as (ax+b)(cx+d), $(a x + b) (c x + d),$ we use FOIL and combine like terms to get Ax2+Bx+C=acx2+(ad+bc)x+bd. $A x^{2} + B x + C = a c x^{2} + (a d + b c) x + b d .$ Notice that AC=acbd=(ad)(bc) $A C = a c b d = (a d) (b c)$ and B=ad+bc $B = a d + b c$ . We can factor Ax2+Bx+C $A x^{2} + B x + C$ if we can find integer factors of the product AC whose sum is B. We illustrate a procedure for doing this in the next example.

Example 10 Factoring Using FOIL and Grouping

Factor.

6x2+17x+7 $6 x^{2} + 17 x + 7$
20x2−7x−6 $20 x^{2} - 7 x - 6$

Solution

Ax2+Bx+C=6x2+17x+7. $A x^{2} + B x + C = 6 x^{2} + 17 x + 7 .$ We have A⋅C=6⋅7=42. $A \cdot C = 6 \cdot 7 = 42 .$ We are looking for two factors of 42 whose sum is 17. We have 42=3⋅14 $42 = 3 \cdot 14$ and 3+14=17. $3 + 14 = 17 .$ So

$6 x 2 + 17 x + 7 = = = = 6 x 2 + 3 x + 14 x + 7 (6 x 2 + 3 x) + (14 x + 7) 3 x (2 x + 1) + 7 (2 x + 1) (3 x + 7) (2 x + 1) 17 x = 3 x + 14 x Group terms . Factor out the common factor . Distributive property$ $\begin{array}{rcll} 6 x^{2} + 17 x + 7 & = & 6 x^{2} + 3 x + 14 x + 7 & 17 x = 3 x + 14 x \\ = & (6 x^{2} + 3 x) + (14 x + 7) & Group terms . \\ = & 3 x (2 x + 1) + 7 (2 x + 1) & Factor out the common factor . \\ = & (3 x + 7) (2 x + 1) & Distributive property \end{array}$
20x2−7x−6. $20 x^{2} - 7 x - 6 .$ We have (20)(−6)=−120. $(20) (- 6) = - 120 .$ The two factors of −120 $- 120$ whose sum is −7 $- 7$ are −15 $- 15$ and 8. So

$20 x 2 - 7 x - 6 = = = = 20 x 2 - 15 x + 8 x - 6 (20 x 2 - 15 x) + (8 x - 6) 5 x (4 x - 3) + 2 (4 x - 3) (5 x + 2) (4 x - 3) - 7 x = - 15 x + 8 x Group terms . Factor out the common factor . Distributive property$ $\begin{array}{rcll} 20 x^{2} - 7 x - 6 & = & 20 x^{2} - 15 x + 8 x - 6 & - 7 x = - 15 x + 8 x \\ = & (20 x^{2} - 15 x) + (8 x - 6) & Group terms . \\ = & 5 x (4 x - 3) + 2 (4 x - 3) & Factor out the common factor . \\ = & (5 x + 2) (4 x - 3) & Distributive property \end{array}$

Practice Problem 10

Factor.
1. 5x2+11x+2 $5 x^{2} + 11 x + 2$
2. 4x2+x−3 $4 x^{2} + x - 3$

Section P.4 Exercises

Concepts and Vocabulary

The polynomials x+2 $x + 2$ and x−2 $x - 2$ are called of the polynomial x2−4. $x^{2} - 4 .$
The polynomial 3y is the factor of the polynomial 3y2+6y. $3 y^{2} + 6 y .$
The GCF of the polynomial 10x3+30x2 $10 x^{3} + 30 x^{2}$ is .
A polynomial that cannot be factored as a product of two polynomials (excluding the constant polynomials ±1 $\pm 1$ ) is said to be .
True or False. When factoring a polynomial whose leading coefficient is positive, we need to consider only its positive factors.
True or False. The polynomial x2+7 $x^{2} + 7$ is irreducible.
True or False. The polynomial x2−4 $x^{2} - 4$ is irreducible.
True or False. Factoring is the opposite of multiplying.

Building Skills

In Exercises 9–26, factor each polynomial by factoring out the GCF.

8x−24 $8 x - 24$
5x+25 $5 x + 25$
−6x2+12x $- 6 x^{2} + 12 x$
−3x2+21 $- 3 x^{2} + 21$
7x2+14x3 $7 x^{2} + 14 x^{3}$
9x3−18x4 $9 x^{3} - 18 x^{4}$
x4+2x3+x2 $x^{4} + 2 x^{3} + x^{2}$
x4−5x3+7x2 $x^{4} - 5 x^{3} + 7 x^{2}$
3x3−x2 $3 x^{3} - x^{2}$
2x3+2x2 $2 x^{3} + 2 x^{2}$
8ax3+4ax2 $8 a x^{3} + 4 a x^{2}$
ax4−2ax2+ax $a x^{4} - 2 a x^{2} + a x$
x(x−y)+3(x−y) $x (x - y) + 3 (x - y)$
2a(a+1)−3(a+1) $2 a (a + 1) - 3 (a + 1)$
3x(2x+y)+5y(2x+y) $3 x (2 x + y) + 5 y (2 x + y)$
2x(5x−3y)+7y(5x−3y) $2 x (5 x - 3 y) + 7 y (5 x - 3 y)$
(y+2)2−3(y+2) ${(y + 2)}^{2} - 3 (y + 2)$
(y+1)2+(y+1) ${(y + 1)}^{2} + (y + 1)$

In Exercises 27–36, factor by grouping.

x3+3x2+x+3 $x^{3} + 3 x^{2} + x + 3$
x3+5x2+x+5 $x^{3} + 5 x^{2} + x + 5$
x3−5x2+x−5 $x^{3} - 5 x^{2} + x - 5$
x3−7x2+x−7 $x^{3} - 7 x^{2} + x - 7$
6x3+4x2+3x+2 $6 x^{3} + 4 x^{2} + 3 x + 2$
3x3+6x2+x+2 $3 x^{3} + 6 x^{2} + x + 2$
12x7+4x5+3x4+x2 $12 x^{7} + 4 x^{5} + 3 x^{4} + x^{2}$
3x7+3x5+x4+x2 $3 x^{7} + 3 x^{5} + x^{4} + x^{2}$
xy+ab+bx+ay $x y + a b + b x + a y$
2a2x−b−bx+2a2 $2 a^{2} x - b - b x + 2 a^{2}$

In Exercises 37–56, factor each trinomial or state that it is irreducible.

x2+7x+12 $x^{2} + 7 x + 12$
x2+8x+15 $x^{2} + 8 x + 15$
x2−6x+8 $x^{2} - 6 x + 8$
x2−9x+14 $x^{2} - 9 x + 14$
x2−3x−4 $x^{2} - 3 x - 4$
x2−5x−6 $x^{2} - 5 x - 6$
x2−4x+13 $x^{2} - 4 x + 13$
x2−2x+5 $x^{2} - 2 x + 5$
2x2+x−36 $2 x^{2} + x - 36$
2x2+3x−27 $2 x^{2} + 3 x - 27$
6x2+17x+12 $6 x^{2} + 17 x + 12$
8x2−10x−3 $8 x^{2} - 10 x - 3$
3x2−11x−4 $3 x^{2} - 11 x - 4$
5x2+7x+2 $5 x^{2} + 7 x + 2$
6x2−3x+4 $6 x^{2} - 3 x + 4$
4x2−4x+3 $4 x^{2} - 4 x + 3$
x2−xy−20y2 $x^{2} - x y - 20 y^{2}$
8p2−2pq−15q2 $8 p^{2} - 2 p q - 15 q^{2}$
15x2−28+x $15 x^{2} - 28 + x$
24y+35+4y2 $24 y + 35 + 4 y^{2}$

In Exercises 57–64, factor each polynomial as a perfect square.

x2+6x+9 $x^{2} + 6 x + 9$
x2+8x+16 $x^{2} + 8 x + 16$
9x2+6x+1 $9 x^{2} + 6 x + 1$
36x2+12x+1 $36 x^{2} + 12 x + 1$
25x2−20x+4 $25 x^{2} - 20 x + 4$
64x2+32x+4 $64 x^{2} + 32 x + 4$
49x2+42x+9 $49 x^{2} + 42 x + 9$
9x2+24x+16 $9 x^{2} + 24 x + 16$

In Exercises 65–74, factor each polynomial using the difference-of-squares pattern.

x2−64 $x^{2} - 64$
x2−121 $x^{2} - 121$
4x2−1 $4 x^{2} - 1$
9x2−1 $9 x^{2} - 1$
16x2−9 $16 x^{2} - 9$
25x2−49 $25 x^{2} - 49$
x4−1 $x^{4} - 1$
x4−81 $x^{4} - 81$
20x4−5 $20 x^{4} - 5$
12x4−75 $12 x^{4} - 75$

In Exercises 75–84, factor each polynomial using the sum-or-difference-of-cubes pattern.

x3+64 $x^{3} + 64$
x3+125 $x^{3} + 125$
x3−27 $x^{3} - 27$
x3−216 $x^{3} - 216$
8−x3 $8 - x^{3}$
27−x3 $27 - x^{3}$
8x3−27 $8 x^{3} - 27$
8x3−125 $8 x^{3} - 125$
40x3+5 $40 x^{3} + 5$
7x3+56 $7 x^{3} + 56$

In Exercises 85–92, factor by making a perfect square.

x4+x2+25 $x^{4} + x^{2} + 25$
x4+x2+1 $x^{4} + x^{2} + 1$
x4+15x2+64 $x^{4} + 15 x^{2} + 64$
x4−7x2+9 $x^{4} - 7 x^{2} + 9$
x4−x2+16 $x^{4} - x^{2} + 16$
x4−16x2+36 $x^{4} - 16 x^{2} + 36$
x4+4 $x^{4} + 4$
x4+64 $x^{4} + 64$

In Exercises 93–124, factor each polynomial completely. If a polynomial cannot be factored, state that it is irreducible.

1−16x2 $1 - 16 x^{2}$
4−25x2 $4 - 25 x^{2}$
x2−6x+9 $x^{2} - 6 x + 9$
x2−8x+16 $x^{2} - 8 x + 16$
4x2+4x+1 $4 x^{2} + 4 x + 1$
16x2+8x+1 $16 x^{2} + 8 x + 1$
2x2−8x−10 $2 x^{2} - 8 x - 10$
5x2−10x−40 $5 x^{2} - 10 x - 40$
2x2+3x−20 $2 x^{2} + 3 x - 20$
2x2−7x−30 $2 x^{2} - 7 x - 30$
x2−24x+36 $x^{2} - 24 x + 36$
x2−20x+25 $x^{2} - 20 x + 25$
3x5+12x4+12x3 $3 x^{5} + 12 x^{4} + 12 x^{3}$
2x5+16x4+32x3 $2 x^{5} + 16 x^{4} + 32 x^{3}$
9x2−1 $9 x^{2} - 1$
16x2−25 $16 x^{2} - 25$
16x2+24x+9 $16 x^{2} + 24 x + 9$
4x2+20x+25 $4 x^{2} + 20 x + 25$
x2+15 $x^{2} + 15$
x2+24 $x^{2} + 24$
45x3+8x2−4x $45 x^{3} + 8 x^{2} - 4 x$
25x3+40x2−9x $25 x^{3} + 40 x^{2} - 9 x$
ax2−7a2x−8a3 $a x^{2} - 7 a^{2} x - 8 a^{3}$
ax2−10a2x−24a3 $a x^{2} - 10 a^{2} x - 24 a^{3}$
x2−16a2+8x+16 $x^{2} - 16 a^{2} + 8 x + 16$
x2−36a2+6x+9 $x^{2} - 36 a^{2} + 6 x + 9$
3x5+12x4y+12x3y2 $3 x^{5} + 12 x^{4} y + 12 x^{3} y^{2}$
4x5+24x4y+36x3y2 $4 x^{5} + 24 x^{4} y + 36 x^{3} y^{2}$
x2−25a2+4x+4 $x^{2} - 25 a^{2} + 4 x + 4$
x2−16a2+4x+4 $x^{2} - 16 a^{2} + 4 x + 4$
18x6+12x5y+2x4y2 $18 x^{6} + 12 x^{5} y + 2 x^{4} y^{2}$
12x6+12x5y+3x4y2 $12 x^{6} + 12 x^{5} y + 3 x^{4} y^{2}$

Applying the Concepts

Garden area. A rectangular garden must have a perimeter of 16 feet. Write a completely factored polynomial whose value gives the area of the garden, assuming that one side is x feet.
Serving tray dimensions. A rectangular serving tray must have a perimeter of 42 inches. Write a completely factored polynomial whose value gives the area of the tray, assuming that one side is x inches.
Box construction. A box with an open top is to be constructed from a rectangular piece of cardboard with length 36 inches and width 16 inches by cutting out squares of equal side length x at each corner and then folding up the sides, as shown in the accompanying figure. Write a completely factored polynomial whose value gives the volume of the box (volume=length×width×height). $(volume = length \times width \times height) .$
Surface area. For the box described in Exercises 127, write a completely factored polynomial whose value gives its surface area (surface area = $=$ sum of the areas of the five sides).
Area of a washer. As shown in the figure, a washer is made by removing a circular disk from the center of another circular disk of radius 2 cm. Assuming that the disk that is removed has radius x, write a completely factored polynomial whose value gives the area of the washer.
Insulation. Insulation must be put in the space between two concentric cylinders. Write a completely factored polynomial whose value gives the volume between the inner and outer cylinders if the inner cylinder has radius 3 feet, the outer cylinder has radius x feet, and both cylinders are 8 feet high.
Grazing pens. A rancher has 2800 feet of fencing that will fence off a rectangular grazing pen for cattle. One edge of the pen borders a straight river and needs no fence. Assuming that the edges of the pen perpendicular to the river are x feet long, write a completely factored polynomial whose value gives the area of the pen.
Centerpiece perimeter. A decorative glass centerpiece for a table has the shape of a rectangle with a semicircular section attached at each end, as shown in the figure. The perimeter of the figure is 48 inches. Write a completely factored polynomial whose value gives the area of the centerpiece.

Beyond the Basics

In Exercises 133–140, factor each expression.

x3+y3+x+y $x^{3} + y^{3} + x + y$
x4−11x2+1 $x^{4} - 11 x^{2} + 1$ [Hint: x4−11x2+1=(x2−1)2−9x2 $x^{4} - 11 x^{2} + 1 = {(x^{2} - 1)}^{2} - 9 x^{2}$ ]
x4+y4+x2y2 $x^{4} + y^{4} + x^{2} y^{2}$
x4+4y4 $x^{4} + 4 y^{4}$
x6−64 $x^{6} - 64$
x6−26x3−27 $x^{6} - 26 x^{3} - 27$
(x2−3x)2−38(x2−3x)−80 ${(x^{2} - 3 x)}^{2} - 38 (x^{2} - 3 x) - 80$
4x2−y2−z2+2yz $4 x^{2} - y^{2} - z^{2} + 2 y z$
In Exercises 110 of Section P.3, page 40, you were asked to show that if a+b+c=0, $a + b + c = 0,$ then a3+b3+c3=3abc. $a^{3} + b^{3} + c^{3} = 3 a b c .$ Now show that: If a3+b3+c3=3abc, $a^{3} + b^{3} + c^{3} = 3 a b c,$ then show that either a+b+c=0 $a + b + c = 0$ or a=b=c. $a = b = c .$ [Hint: From Exercises 110, page 40, we have a3+b3+c3−3abc= $a^{3} + b^{3} + c^{3} - 3 a b c =$

(a+b+c)(a2+b2+c2−ab−bc−ca) $(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

=12(a+b+c)[(a−b)2+(b−c)2+(c−a)2)] $= \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2})]$ .
Factor (x−y)3+(y−z)3+(z−x)3. ${(x - y)}^{3} + {(y - z)}^{3} + {(z - x)}^{3} .$
Factor z3(x−y)3+x3(y−z)3+y3(z−x)3. $z^{3} {(x - y)}^{3} + x^{3} {(y - z)}^{3} + y^{3} {(z - x)}^{3} .$
Simplify (x2−y2)3+(y2−z2)3+(z2−x2)3(x−y)3+(y−z)3+(z−x)3. $\frac{{(x^{2} - y^{2})}^{3} + {(y^{2} - z^{2})}^{3} + {(z^{2} - x^{2})}^{3}}{{(x - y)}^{3} + {(y - z)}^{3} + {(z - x)}^{3}} .$

Getting Ready for the Next Section

In Exercises 145–148, combine each expression to form a single fraction in lowest terms.

14+16 $\frac{1}{4} + \frac{1}{6}$
38−112 $\frac{3}{8} - \frac{1}{12}$
920⋅1627 $\frac{9}{20} \cdot \frac{16}{27}$
23⋅56 $\frac{2}{3} \cdot \frac{5}{6}$

In Exercises 149–152, factor the numerator and denominator and write each expression in lowest terms.

2x4x+6 $\frac{2 x}{4 x + 6}$
2x3+3xx2+x $\frac{2 x^{3} + 3 x}{x^{2} + x}$
x2+3x+2x2+4x+3 $\frac{x^{2} + 3 x + 2}{x^{2} + 4 x + 3}$
x2−x−6x2−4 $\frac{x^{2} - x - 6}{x^{2} - 4}$

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Table of Contents for Section P.4 Factoring Polynomials

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Table of Contents for
Section P.4 Factoring Polynomials