2-4 Zeros Of Polynomial Functions 2x2v1g

2-4 Zeros of Polynomial Functions List all possible rational zeros of each function. Then determine which, if any, are zeros.

1. g(x) = x4 – 6x3 – 31x2 + 216x − 180 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −180. Therefore, the possible rational zeros of g are . By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero.

Because (x − 1) and (x − 5) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g 2 (x) = (x − 1)(x − 5)(x − 36). Factoring the quadratic expression yields f (x) = (x – 1)(x – 5)(x – 6)(x + 6). Thus, the rational zeros of g are 1, 5, 6, and −6.

2. f (x) = 4x3 – 24x2 – x + 6 SOLUTION:   or

The leading coefficient is 4 and the constant term is 6. The possible rational zeros are . By using synthetic division, it can be determined that x = 6 is a rational zero.

2

Because (x − 6) is a factor of f (x), we can write a factored form of f (x) as f (x) = (x − 6)(4x − 1). To find zeros of 2

4x − 1, we can set it equal to zero and solve for x.

Thus, the rational zeros of f are

3. g(x) = x4 – x3 – 31x2 + x + 30 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 30. Therefore, the possible rational zeros of g are . eSolutions Manual - Powered by Cognero By using synthetic division, it can be determined that x = 1 is a rational zero.

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2-4 Thus, Zeros Polynomial Functions theof rational zeros of f are 3. g(x) = x4 – x3 – 31x2 + x + 30 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 30. Therefore, the possible rational zeros of g are . By using synthetic division, it can be determined that x = 1 is a rational zero.


Because (x − 1) and (x − 6) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g 2

(x) = (x − 1)(x − 6)(x +6x + 5). Factoring the quadratic expression yields f (x) = (x – 1)(x – 6)(x + 5)(x + 1). Thus, the rational zeros of g are 1, 6, −5, and −1.

4. g(x) = −4x4 + 35x3 – 87x2 + 56x + 20 SOLUTION:   The leading coefficient is −4 and the constant term is 20. The possible rational zeros are

or

. By using synthetic division, it can be determined that x = 2 is a rational zero.



(x) = (x − 2)(x − 5)(−4x +7x + 2). Factoring the quadratic expression yields f (x) = (x – 2)(x – 5)(x − 2)(−4x − 1) or 2

(x − 2) (x − 5)(−4x − 1). Thus, the rational zeros of g are

5. h(x) = 6x4 + 13x3 – 67x2 – 156x – 60 SOLUTION:   The leading coefficient is 6 and the constant term is −60. The possible rational zeros are or

. eSolutions Manual - Powered by Cognero

By using synthetic division, it can be determined that

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is a rational zero.


(x) = (x − 2)(x − 5)(−4x +7x + 2). Factoring the quadratic expression yields f (x) = (x – 2)(x – 5)(x − 2)(−4x − 1) or 2 2-4 (x Zeros Functions − 2) (xof− Polynomial 5)(−4x − 1). Thus, the rational zeros of g are

5. h(x) = 6x4 + 13x3 – 67x2 – 156x – 60 SOLUTION:   The leading coefficient is 6 and the constant term is −60. The possible rational zeros are or

.

is a rational zero.


By using synthetic division on the depressed polynomial, it can be determined that

Because

and

is a rational zero.

are factors of h(x), we can use the final quotient to write a factored form of h(x) as Factoring the quadratic expression yields 2

Because the factor (x − 12) yields no rational zeros, the rational zeros of h are

6. f (x) = 18x4 + 12x3 + 56x2 + 48x − 64 SOLUTION:   The leading coefficient is 18 and the constant term is −64. The possible rational zeros are

or

By using synthetic division, it can be determined that eSolutions Manual - Powered by Cognero

.

is a rational zero. Page 3

2

Because the factor (x − 12) yields no rational zeros, the rational zeros of h are

2-4 Zeros of Polynomial Functions 6. f (x) = 18x4 + 12x3 + 56x2 + 48x − 64 SOLUTION:   The leading coefficient is 18 and the constant term is −64. The possible rational zeros are

or

.

is a rational zero.



Because

and

is a rational zero.

are factors of f (x), we can use the final quotient to write a factored form of f (x) as Factoring the quadratic expression yields 2

Because the factor (x + 4) yields no real zeros, the rational zeros of f are

7. h(x) = x5 – 11x4 + 49x3 – 147x2 + 360x – 432 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −432. Therefore, the possible rational zeros of g are . By using synthetic division, it can be determined that x = 3 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero. eSolutions Manual - Powered by Cognero

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By using synthetic division on the new depressed polynomial, it can be determined that x = 4 is a repeated rational

2

Because the factor (x + 4) yields no real zeros, the rational zeros of f are

2-4 Zeros of Polynomial Functions 7. h(x) = x5 – 11x4 + 49x3 – 147x2 + 360x – 432 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −432. Therefore, the possible rational zeros of g are . By using synthetic division, it can be determined that x = 3 is a rational zero.


By using synthetic division on the new depressed polynomial, it can be determined that x = 4 is a repeated rational zero.

2

Because (x − 3) and (x − 4) are factors of h(x), we can use the final quotient to write a factored form of h(x) as h 2 2 2 (x) = (x − 3)(x − 4) (x + 9). Because the factor (x + 9) yields no real zeros, the rational zeros of h are 3 and 4 (multiplicity: 2).

8. g(x) = 8x5 + 18x4 – 5x3 – 72x2 – 162x + 45 SOLUTION:   or

The leading coefficient is 8 and the constant term is 45. The possible rational zeros are

.

. By using synthetic division, it can be determined that

is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x =

eSolutions Manual - Powered by Cognero

is a rational zero.

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2

2-4

Because (x − 3) and (x − 4) are factors of h(x), we can use the final quotient to write a factored form of h(x) as h 2 2 2 (x) = (x −of3)(x − 4) (x + 9).Functions Because the factor (x + 9) yields no real zeros, the rational zeros of h are 3 and 4 Zeros Polynomial (multiplicity: 2).

8. g(x) = 8x5 + 18x4 – 5x3 – 72x2 – 162x + 45 SOLUTION:   or

The leading coefficient is 8 and the constant term is 45. The possible rational zeros are

.

is a rational zero.

. By using synthetic division, it can be determined that


Because

and

is a rational zero.

are factors of g(x), we can use the final quotient to write a factored form of f (x) as 3

To find zeros of 8x − 72, we can set it equal to zero and solve for x.

3

Because the factor (8x −72) yields no rational zeros, the rational zeros of g are

9. MANUFACTURING  The specifications for the dimensions of a new cardboard container are shown. If the 3

2

volume of the container is modeled by V(h) = 2h – 9h + 4h and it will hold 45 cubic inches of merchandise, what are the container's dimensions?

SOLUTION:


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3

2

Substitute V(h) = 45 into V(h) = 2h – 9h + 4h and apply the Rational Zeros Theorem to find possible rational zeros of the function.

3 2-4 Because Zeros of thePolynomial factor (8x −72)Functions yields no rational zeros, the rational zeros of g are

9. MANUFACTURING  The specifications for the dimensions of a new cardboard container are shown. If the 3

2

volume of the container is modeled by V(h) = 2h – 9h + 4h and it will hold 45 cubic inches of merchandise, what are the container's dimensions?

SOLUTION:   3

2

Substitute V(h) = 45 into V(h) = 2h – 9h + 4h and apply the Rational Zeros Theorem to find possible rational zeros of the function.

The leading coefficient is 2 and the constant term is −45. The possible rational zeros are

or

. By using synthetic division, it can be determined that h = 5 is a rational zero.

2

The depressed polynomial 2x + x + 9 has no real zeros. Thus, h = 5. The dimensions of the container are 5, 5 −4 or 1, and 2(5) − 1 or 9. Solve each equation.

10. x4 + 2x3 – 7x2 – 20x – 12 = 0 SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −12. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, and ±12. By using synthetic division, it can be determined that x = −1 is a rational zero.


Because (x + 1) and (x − 3) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 2 (x + 1)(x − 3)(x + 4x + 4). Factoring the quadratic expression yields 0 = (x + 1)(x − 3)(x + 2) . Thus, the solutions are −1, 3, and −2 (multiplicity: 2).

11. x4 + 9x3 + 23x2 + 3x − 36 = 0


Page 7

SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is

2-4

Because (x + 1) and (x − 3) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 2 (x + 1)(x − 3)(x + 4x + 4). Factoring the quadratic expression yields 0 = (x + 1)(x − 3)(x + 2) . Thus, the solutions Zeros Polynomial are −1, 3,of and −2 (multiplicity:Functions 2).

11. x4 + 9x3 + 23x2 + 3x − 36 = 0 SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −36. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = −4 is a rational zero.

Because (x − 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 2 (x − 1)(x + 4)(x + 9x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x + 4)(x + 3) . Thus, the solutions are 1, −4, and −3 (multiplicity: 2).

12. x4 – 2x3 – 7x2 + 8x + 12 = 0 SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 12. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, and ±12. By using synthetic division, it can be determined that x = −1 is a rational zero.


Because (x + 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 (x + 1)(x − 2)(x − x − 6). Factoring the quadratic expression yields 0 = (x + 1)(x − 2)(x − 3)(x + 2). Thus, the solutions are −1, 2, 3, and −2.

13. x4 – 3x3 – 20x2 + 84x – 80 = 0 SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −80. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that x = 4 is a rational zero.



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2-4

Because (x + 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 (x + 1)(xof − 2)(x − x − 6). Factoring the quadratic expression yields 0 = (x + 1)(x − 2)(x − 3)(x + 2). Thus, the Zeros Polynomial Functions solutions are −1, 2, 3, and −2.

13. x4 – 3x3 – 20x2 + 84x – 80 = 0 SOLUTION:   Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −80. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that x = 4 is a rational zero.


Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2 2 (x − 4)(x + 5)(x − 4x + 4). Factoring the quadratic expression yields 0 = (x − 4)(x + 5)(x − 2) . Thus, the solutions are 4, −5, and 2 (multiplicity: 2).

14. x4 + 34x = 6x3 + 21x2 – 48 SOLUTION:   4

3

2

The equation can be written as x − 6x − 21x + 34x + 48 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 48. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that x = −1 is a rational zero.


Because (x + 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2

(x + 1)(x − 2)(x − 5x − 24). Factoring the quadratic expression yields 0 = (x + 1)(x − 2)(x + 3)(x − 8). Thus, the solutions are −1, 2, −3, and 8.

15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26 SOLUTION:   4

3

2

The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational zeros are


or

By using synthetic division, it can be determined that x =

.

is a rational zero.

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Because (x + 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 = 2

+ 1)(xof − 2)(x − 5x − 24). Functions Factoring the quadratic expression yields 0 = (x + 1)(x − 2)(x + 3)(x − 8). Thus, the 2-4 (x Zeros Polynomial solutions are −1, 2, −3, and 8.

15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26 SOLUTION:   4

3

2

The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational or

zeros are

.


is a rational zero.


Because

and

is a rational zero.

are factors of the equation, we can use the final quotient to write a factored form as . Factoring the quadratic expression yields

the solutions are

,

.Thus,

, and −4 (multiplicity: 2).

16. –12x4 + 77x3 = 136x2 – 33x – 18 SOLUTION:   4

3

2

The equation can be written as −12x + 77x − 136x + 33x + 18 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational zeros are

or


.

is a rational zero.



is a rational zero.

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. Factoring the quadratic expression yields

.Thus,

2-4 the Zeros of Polynomial solutions are , , and Functions −4 (multiplicity: 2). 16. –12x4 + 77x3 = 136x2 – 33x – 18 SOLUTION:   4

3

2

The equation can be written as −12x + 77x − 136x + 33x + 18 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational or

zeros are

.


is a rational zero.


and

Because

is a rational zero.


.

Thus, the solutions are

17. SALES  The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 3

2

2x – 2x + 4x, where x is the number of days after the first day of the month. How many days will it take the store to make $16,000?

SOLUTION:   3

2

Substitute S(x) = 16 into S(x) = 2x – 2x + 4x and apply the Rational Zeros Theorem to find possible rational zeros of the function.

3

2

The equation can be written as 2(x − x + 2x − 8) = 0. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros are ±1, ±2, ±4, and ±8. By using synthetic division, it can be determined that x = 2 is a rational zero.

2

The depressed polynomial x + x + 4 has no real zeros. Thus, x = 2. The store will make $16,000 in 2 days. eSolutions Manual - Powered by Cognero

Page 11

Determine an interval in which all real zeros of each function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. 4

3

2


.

2-4 Thus, Zeros Polynomial Functions theof solutions are 17. SALES  The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 3

2

2x – 2x + 4x, where x is the number of days after the first day of the month. How many days will it take the store to make $16,000?

SOLUTION:   3

2

Substitute S(x) = 16 into S(x) = 2x – 2x + 4x and apply the Rational Zeros Theorem to find possible rational zeros of the function.

3

2

The equation can be written as 2(x − x + 2x − 8) = 0. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros are ±1, ±2, ±4, and ±8. By using synthetic division, it can be determined that x = 2 is a rational zero.

2

The depressed polynomial x + x + 4 has no real zeros. Thus, x = 2. The store will make $16,000 in 2 days. Determine an interval in which all real zeros of each function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. 18. f (x) = x4 – 9x3 + 12x2 + 44x – 48

SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 10].

Test a lower bound of c = −3 and an upper bound of c = 10.

Every number in the last line is alternately nonnegative and nonpositive, so −3 is a lower bound.

Every number in the last line is nonnegative, so 10 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48. Because the real zeros are in the interval [−3, 10], narrow this list to just ±1, ±2, ±3, 4, 6, and 8. From the graph, it appears that −2, 1, 4, and 6 are reasonable. By using synthetic division, it can be determined that x = −2 is a rational zero. eSolutions Manual - Powered by Cognero


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2-4 Zeros of Polynomial2Functions

The depressed polynomial x + x + 4 has no real zeros. Thus, x = 2. The store will make $16,000 in 2 days.

Determine an interval in which all real zeros of each function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. 18. f (x) = x4 – 9x3 + 12x2 + 44x – 48

SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 10].



Every number in the last line is nonnegative, so 10 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48. Because the real zeros are in the interval [−3, 10], narrow this list to just ±1, ±2, ±3, 4, 6, and 8. From the graph, it appears that −2, 1, 4, and 6 are reasonable. By using synthetic division, it can be determined that x = −2 is a rational zero.


Because (x + 2) and (x − 1) are factors of the equation, we can use the final quotient to write a factored form as f 2

(x) = (x + 2)(x − 1)(x − 10x + 24). Factoring the quadratic expression yields f (x) = (x + 2)(x − 1)(x − 4)(x − 6). Thus, the solutions are −2, 1, 4, and 6.

19. f (x) = 2x4 – x3 – 29x2 + 34x + 24 SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−6, 5].


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Because (x + 2) and (x − 1) are factors of the equation, we can use the final quotient to write a factored form as f 2

2-4

(x) = (x + 2)(x − 1)(x − 10x + 24). Factoring the quadratic expression yields f (x) = (x + 2)(x − 1)(x − 4)(x − 6). Zeros Polynomial Thus, theof solutions are −2, 1, Functions 4, and 6.

19. f (x) = 2x4 – x3 – 29x2 + 34x + 24 SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−6, 5].



Every number in the last line is nonnegative, so 5 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the or ±1, ±2, ±3, ±4, ±6, ±8, ±12,

constant term is 24. The possible rational zeros are ±24,

, and

. Because the real zeros are in the interval [−6, 5], narrow this list to just ±1, ±2, ±3, ±4, −6,

and

. From the graph, it appears that

,

are reasonable.

By using synthetic division, it can be determined that x = −4 is a rational zero.


Because (x + 4) and

is a rational zero.

are factors of the equation, we can use the final quotient to write a factored form as . Factoring the quadratic expression yields . Thus, the solutions are

eSolutions Manual4- Powered 3 by Cognero 2

20. g(x) = 2x + 4x – 18x – 4x + 16 SOLUTION:

Page 14


2-4 Zeros of Polynomial Functions . Thus, the solutions are 20. g(x) = 2x4 + 4x3 – 18x2 – 4x + 16 SOLUTION:   Graph g(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−6, 4].



Every number in the last line is nonnegative, so 4 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is 16. The possible rational zeros are

or ±1, ±2, ±4, ±8, ±16, and ±

real zeros are in the interval [−6, 4], narrow this list to just ±1, ±2, ±3, ±4, and

. Because the

. From the graph, it appears that

−4, −1, 1, and 2 are reasonable. By using synthetic division, it can be determined that x = −4 is a rational zero.


Because (x + 4) and (x + 1) are factors of the equation, we can use the final quotient to write a factored form as g 2 (x) = (x + 4)(x + 1)(2x − 6x + 4). Factoring the quadratic expression yields g(x) = 2(x + 4)(x + 1)(x − 2)(x − 1). Thus, the solutions are −4, −1, 2, and 1.

21. g(x) = 6x4 – 33x3 – 6x2 + 123x – 90 SOLUTION:   Graph g(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−4, 7].


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2-4

Because (x + 4) and (x + 1) are factors of the equation, we can use the final quotient to write a factored form as g 2 (x) = (x +of4)(x + 1)(2x − 6x Functions + 4). Factoring the quadratic expression yields g(x) = 2(x + 4)(x + 1)(x − 2)(x − 1). Zeros Polynomial Thus, the solutions are −4, −1, 2, and 1.

21. g(x) = 6x4 – 33x3 – 6x2 + 123x – 90 SOLUTION:   Graph g(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−4, 7].



Every number in the last line is nonnegative, so 7 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 90. The possible rational zeros are or

.

Because the real zeros are in the interval [−4, 7], narrow this list to just . From the graph, it appears that −2, 1,

, and 5 are

reasonable. By using synthetic division, it can be determined that x = −2 is a rational zero.


Because (x + 2) and (x − 1) are factors of the equation, we can use the final quotient to write a factored form as g 2 (x) = (x + 2)(x − 1)(6x − 39x + 45). Factoring the quadratic expression yields g(x) = 3(x + 2)(x − 1)(2x − 3)(x − 5). Thus, the solutions are −2, 1,

, and 5.

22. f (x) = 2x4 – 17x3 + 39x2 – 16x – 20 eSolutions Manual - Powered by Cognero SOLUTION:

Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−2, 9].

Page 16

(x) = (x + 2)(x − 1)(6x − 39x + 45). Factoring the quadratic expression yields g(x) = 3(x + 2)(x − 1)(2x − 3)(x − 5). Thus, the solutions are −2, 1,

, and 5.

2-4 Zeros of Polynomial Functions 22. f (x) = 2x4 – 17x3 + 39x2 – 16x – 20 SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−2, 9].



Every number in the last line is nonnegative, so 9 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is −20. The possible rational zeros are ±

or ±1, ±2, ±4, ±5, ±10, ±20, ±

. Because the real zeros are in the interval [−2, 9], narrow this list to just ±1, ±2, 4, 5, ±

graph, it appears that

, and

, and

. From the

are reasonable.


is a rational zero.


Because

and (x − 5) are factors of the equation, we can use the final quotient to write a factored form as . Factoring the quadratic expression yields

Thus, the solutions are

.

.

23. f (x) = 2x4 – 13x3 + 21x2 + 9x – 27 eSolutions Manual - Powered by Cognero

SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the

Page 17


2-4 Thus, Zeros Polynomial Functions theof solutions are

.

.

23. f (x) = 2x4 – 13x3 + 21x2 + 9x – 27 SOLUTION:   Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−2, 7].



Every number in the last line is nonnegative, so 7 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is −27. The possible rational zeros are

or ±1, ±3, ±9, ±27, ±

Because the real zeros are in the interval [−2, 7], narrow this list to just ±1, 3, ± appears that −1 and

,±

,±

, and

,±

and ±

.

. From the graph, it

are reasonable.



Because (x + 1) and


Thus, the solutions are −1, eSolutions Manual by 5 - Powered 4 3 Cognero 2

.

, and 3 (multiplicity: 2).

24. h(x) = x – x – 9x + 5x + 16x – 12 SOLUTION:

is a rational zero.

Page 18


.

2-4 Thus, Zeros Polynomial theof solutions are −1, Functions , and 3 (multiplicity: 2). 24. h(x) = x5 – x4 – 9x3 + 5x2 + 16x – 12 SOLUTION:   Graph h(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 5].



Every number in the last line is nonnegative, so 5 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −12. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, and ±12. Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±2, ±3, and 4. From the graph, it appears that −2, 1, and 3 are reasonable. By using synthetic division, it can be determined that x = −2 is a rational zero.


By using synthetic division on the new depressed polynomial, it can be determined that x = 3 is a rational zero.

Because (x + 2), (x − 1), and (x − 3) are factors of the equation, we can use the final quotient to write a factored 2 form as h(x) = (x + 2)(x − 1)(x − 3)(x + x − 2). Factoring the quadratic expression yields f (x) = (x + 2)(x − 1)(x − 2

2

3)(x + 2)(x − 1) or (x + 2) (x − 1) (x − 3). Thus, the solutions are 3, −2 (multiplicity: 2), and 1 (multiplicity: 2).

25. h(x) = 4x5 – 20x4 + 5x3 + 80x2 – 75x + 18 SOLUTION:   Graph h(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 5]. eSolutions Manual - Powered by Cognero

Page 19

3)(x + 2)(x − 1) or (x + 2) (x − 1) (x − 3). Thus, the solutions are 3, −2 (multiplicity: 2), and 1 (multiplicity: 2).

25. h(x) = 4x5 – 20x4 + 5x3 + 80x2 – 75x + 18

2-4 SOLUTION:   Zeros of Polynomial Functions Graph h(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 5].



Every number in the last line is nonnegative, so 5 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 4 and the or ±1, ±3, ±6, ±9, ±18, ±

constant term is 18. The possible rational zeros are

,±

±

,±

, and ±

. Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±3, ±

±

,±

, and ±

. From the graph, it appears that −2,

,±

,± ,

, ,

and 3 are reasonable.



is a rational zero.

By using synthetic division on the new depressed polynomial, it can be determined that x = 3 is a rational zero.

Because form as

are factors of the equation, we can use the final quotient to write a factored . Factoring the quadratic expression yields . Thus, the solutions are −2,


and 3 (multiplicity: 2). Describe the possible real zeros of each function.

(multiplicity: 2), Page 20

are factors of the equation, we can use the final quotient to write a factored

Because

2-4 Zeros of Polynomial Functions form as

. Factoring the quadratic expression yields . Thus, the solutions are −2,

(multiplicity: 2),

and 3 (multiplicity: 2). Describe the possible real zeros of each function.

26. f (x) = –2x3 – 3x2 + 4x + 7 SOLUTION:   Examine the variations in sign for f (x) and for f (−x). 3 2 f(x) = –2x – 3x + 4x + 7 f(x) has one variation in sign.

f(−x) has 2 variations in sign. By Descartes’ Rule of Signs, f (x) has 1 positive zero and either 2 or 0 negative zeros.

27. f (x) = 10x4 – 3x3 + 8x2 – 4x – 8 SOLUTION:   Examine the variations in sign for f (x) and for f (−x). 4 3 2 f(x) = 10x – 3x + 8x – 4x – 8 f(x) has three variations in sign.

f(−x) has 1 variation in sign. By Descartes’ Rule of Signs, f (x) has either 3 or 1 positive zeros and 1 negative zero.

28. f (x) = –3x4 – 5x3 + 4x2 + 2x – 6 SOLUTION:   Examine the variations in sign for f (x) and for f (−x). 4 3 2 f(x) = –3x – 5x + 4x + 2x – 6 f(x) has two variations in sign.

f(−x) has two variations in sign. By Descartes’ Rule of Signs, f (x) has either 2 or 0 positive zeros and either 2 or 0 negative zeros.

29. f (x) = 12x4 + 6x3 + 3x2 – 2x + 12 SOLUTION:   Examine the variations in sign for f (x) and for f (−x). 4 3 2 f(x) = 12x + 6x + 3x – 2x + 12 f(x) has two variations in sign.

f(−x)Manual has two variations in sign. eSolutions - Powered by Cognero

By Descartes’ Rule of Signs, f (x) has either 2 or 0 positive zeros and either 2 or 0 negative zeros.

30.

5

4

3

2

Page 21

hasof twoPolynomial variations in sign. 2-4 f(−x) Zeros Functions

By Descartes’ Rule of Signs, f (x) has either 2 or 0 positive zeros and either 2 or 0 negative zeros.

29. f (x) = 12x4 + 6x3 + 3x2 – 2x + 12 SOLUTION:   Examine the variations in sign for f (x) and for f (−x). 4 3 2 f(x) = 12x + 6x + 3x – 2x + 12 f(x) has two variations in sign.

f(−x) has two variations in sign. By Descartes’ Rule of Signs, f (x) has either 2 or 0 positive zeros and either 2 or 0 negative zeros.

30. g(x) = 4x5 + 3x4 + 9x3 – 8x2 + 16x – 24 SOLUTION:   Examine the variations in sign for g(x) and for g(−x). 5 4 3 2 g(x) = 4x + 3x + 9x – 8x + 16x – 24 g(x) has three variations in sign.

g(−x) has two variations in sign. By Descartes’ Rule of Signs, g(x) has either 3 or 1 positive zeros and either 2 or 0 negative zeros.

31. h(x) = –4x5 + x4 – 8x3 – 24x2 + 64x – 124 SOLUTION:   Examine the variations in sign for h(x) and for h(−x). 5 4 3 2 h(x) = –4x + x – 8x – 24x + 64x – 124 h(x) has four variations in sign.

h(−x) has one variation in sign. By Descartes’ Rule of Signs, h(x) has either 4, 2, or 0 positive zeros and 1 negative zero. Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. 32. 3, –4, 6, –1

SOLUTION:   Using the Linear Factorization Theorem and the zeros 3, −4, 6, and −1, write f (x) as follows. f(x) = a[x − (3)][x − (−4)][x − (6)][x − (−1)] Let a = 1. Then write the function in standard form.

4

3

2

Therefore, a function of least degree that has 3, −4, 6, and −1 as zeros is f (x) = x – 4x – 23x + 54x + 72 or any nonzero multiple of f (x). eSolutions Manual - Powered by Cognero

33. –2, –4, –3, 5 SOLUTION:

Page 22

hasof one variation in sign. 2-4 h(−x) Zeros Polynomial Functions

By Descartes’ Rule of Signs, h(x) has either 4, 2, or 0 positive zeros and 1 negative zero.

Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. 32. 3, –4, 6, –1

SOLUTION:   Using the Linear Factorization Theorem and the zeros 3, −4, 6, and −1, write f (x) as follows. f(x) = a[x − (3)][x − (−4)][x − (6)][x − (−1)] Let a = 1. Then write the function in standard form.

4

3

2

Therefore, a function of least degree that has 3, −4, 6, and −1 as zeros is f (x) = x – 4x – 23x + 54x + 72 or any nonzero multiple of f (x).

33. –2, –4, –3, 5 SOLUTION:   Using the Linear Factorization Theorem and the zeros −2, −4, −3, and 5, write f (x) as follows. f(x) = a[x − (−2)][x − (−4)][x − (−3)][x − (5)] Let a = 1. Then write the function in standard form.

4

3

2

Therefore, a function of least degree that has −2, −4, −3, and 5 as zeros is f (x) = x + 4x – 19x – 106x – 120 or any nonzero multiple of f (x).

34. –5, 3, 4 + i SOLUTION:   Because 4 + i is a zero and the polynomial is to have real coefficients, you know that 4 − i must also be a zero. Using the Linear Factorization Theorem and the zeros –5, 3, 4 + i, and 4 − i, write f (x) as follows. f(x) = a[x − (−5)][x − (3)][x − (4 + i)][x − (4 − i)] Let a = 1. Then write the function in standard form.

4

3

2

Therefore, a function of least degree that has –5, 3, 4 + i, and 4 − i as zeros is f (x) = x – 6x – 14x + 154x − 255 or any nonzero multiple of f (x).

35. –1, 8, 6 − i SOLUTION:   Because 6 − i is a zero and the polynomial is to have real coefficients, you know that 6 + i must also be a zero. Using the Linear Factorization Theorem and the zeros –1, 8, 6 − i, and 6 + i, write f (x) as follows. f(x) = a[x − (−1)][x − (8)][x − (6 − i)][x − (6 + i)] Let a = 1. Then write the function in standard form.


Page 23

function of leastFunctions degree that has –5, 3, 4 + i, and 4 − i as zeros is f (x) = x 2-4 Therefore, Zeros ofaPolynomial

4

3

2

– 6x – 14x + 154x − 255

or any nonzero multiple of f (x).

35. –1, 8, 6 − i SOLUTION:   Because 6 − i is a zero and the polynomial is to have real coefficients, you know that 6 + i must also be a zero. Using the Linear Factorization Theorem and the zeros –1, 8, 6 − i, and 6 + i, write f (x) as follows. f(x) = a[x − (−1)][x − (8)][x − (6 − i)][x − (6 + i)] Let a = 1. Then write the function in standard form.

4

3

2

Therefore, a function of least degree that has –1, 8, 6 − i, and 6 + i as zeros is f (x) = x – 19x + 113x – 163x − 296 or any nonzero multiple of f (x).

36. 2

, –2

, –3, 7

SOLUTION:   Using the Linear Factorization Theorem and the zeros 2

, –2

, –3, and 7, write f (x) as follows.

f(x) = a[x − (2 )][x − (−2 )][x − (−3)][x − (7)] Let a = 1. Then write the function in standard form.

Therefore, a function of least degree that has 2 or any nonzero multiple of f (x).

37. −5, 2, 4 −

4

, –2

3

2

, –3, and 7 as zeros is f (x) = x – 4x – 41x + 80x + 420

,4+

SOLUTION:   Using the Linear Factorization Theorem and the zeros −5, 2, 4 − f(x) = a[x − (−5)][x − (2)][x − (4 − )][x − (4 + Let a = 1. Then write the function in standard form.

,–

, write f (x) as follows.

)]

Therefore, a function of least degree that has −5, 2, 4 − − 130 or any nonzero multiple of f (x).

38.

, and 4 +

, and 4 +

4

3

2

as zeros is f (x) = x – 5x – 21x + 119x

, 4i

SOLUTION:   Because 4i is a zero and the polynomial is to have real coefficients, you know that −4i must also be a zero. Using the Linear Factorization Theorem and the zeros ,– , 4i, and −4i, write f (x) as follows. f(x) = a[x − ( )][x − (− )][x − (4i)][x − (−4i)] Let a = 1. Then write the function in standard form.


Page 24

function of leastFunctions degree that has −5, 2, 4 − 2-4 Therefore, Zeros ofaPolynomial

, and 4 +

4

3

2

as zeros is f (x) = x – 5x – 21x + 119x

− 130 or any nonzero multiple of f (x).

38.

,–

, 4i

SOLUTION:   Because 4i is a zero and the polynomial is to have real coefficients, you know that −4i must also be a zero. Using the Linear Factorization Theorem and the zeros ,– , 4i, and −4i, write f (x) as follows. f(x) = a[x − ( )][x − (− )][x − (4i)][x − (−4i)] Let a = 1. Then write the function in standard form.

Therefore, a function of least degree that has nonzero multiple of f (x).

39.

,–

,–

4

2

, 4i, and −4i as zeros is f (x) = x + 9x – 112 or any

, 3 − 4i

SOLUTION:   Because 3 − 4i is a zero and the polynomial is to have real coefficients, you know that 3 + 4i must also be a zero. Using the Linear Factorization Theorem and the zeros ,– , 3 − 4i, and 3 + 4i, write f (x) as follows. f(x) = a[x − ( )][x − (− )][x − (3 − 4i)][x − (3 + 4i)] Let a = 1. Then write the function in standard form.

Therefore, a function of least degree that has − 150 or any nonzero multiple of f (x).

40. 2 +

,2 −

,–

4

3

2

, 3 − 4i, and 3 + 4i as zeros is f (x) = x – 6x + 19x + 36x

, 4 + 5i

SOLUTION:   Because 4 + 5i is a zero and the polynomial is to have real coefficients, you know that 4 − 5i must also be a zero. Using the Linear Factorization Theorem and the zeros 2 + ,2 − , 4 + 5i, and 4 − 5i, write f (x) as follows. f(x) = a[x − (2 + )][x − (2 − )][x − (4 + 5i)][x − (4 − 5i)] Let a = 1. Then write the function in standard form.

Therefore, a function of least degree that has 2 + – 172x + 41 or any nonzero multiple of f (x).

41. 6 −

,6+

,2 −

4

3

, 4 + 5i, and 4 − 5i as zeros is f (x) = x – 12x + 74x

2

, 8 – 3i

SOLUTION:   Because 3i is aby zero and the eSolutions Manual8- – Powered Cognero

polynomial is to have real coefficients, you know that 8 + 3i must also be a zero. Page 25 Using the Linear Factorization Theorem and the zeros 6 − ,6+ , 8 – 3i, and 8 + 3i, write f (x) as follows. f(x) = a[x − (6 −

)][x − (6 +

)][x − (8 – 3i)][x − (8 + 3i)]

function of leastFunctions degree that has 2 + 2-4 Therefore, Zeros ofaPolynomial

,2 −

4

3

, 4 + 5i, and 4 − 5i as zeros is f (x) = x – 12x + 74x

2

– 172x + 41 or any nonzero multiple of f (x).

41. 6 −

,6+

, 8 – 3i

SOLUTION:   Because 8 – 3i is a zero and the polynomial is to have real coefficients, you know that 8 + 3i must also be a zero. Using the Linear Factorization Theorem and the zeros 6 − ,6+ , 8 – 3i, and 8 + 3i, write f (x) as follows. f(x) = a[x − (6 − )][x − (6 + )][x − (8 – 3i)][x − (8 + 3i)] Let a = 1. Then write the function in standard form.

Therefore, a function of least degree that has 6 −

,6+

4 3 , 8 – 3i, and 8 + 3i as zeros is f (x) = x – 28x +

2

296x – 1372x + 2263 or any nonzero multiple of f (x). Write each function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all of its zeros. 42. g(x) = x4 – 3x3 – 12x2 + 20x + 48

SOLUTION:   a. g(x) has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48. By using synthetic division, it can be determined that x = 4 is a rational zero.


2

2

The remaining quadratic factor (x + 4x + 4) can be written as (x + 2) . 2 So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x – 4)(x – 3)(x + 2) 2 b. g(x) written as a product of linear factors is g(x) = (x – 4)(x – 3)(x + 2) . c. The zeros are 4, 3, and –2 (multiplicity: 2).

43. g(x) = x4 – 3x3 – 12x2 + 8 SOLUTION:   a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8. By using synthetic division, it can be determined that x = −2 is a rational zero.



Page 26

2

The remaining quadratic factor (x − 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros.

2-4

The remaining quadratic factor (x + 4x + 4) can be written as (x + 2) . 2 So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x – 4)(x – 3)(x + 2) 2 b. g(x) written as a product of linear factors is g(x) = (x – 4)(x – 3)(x + 2) . Zeros of Polynomial c. The zeros are 4, 3, and –2 Functions (multiplicity: 2).

43. g(x) = x4 – 3x3 – 12x2 + 8 SOLUTION:   a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8. By using synthetic division, it can be determined that x = −2 is a rational zero.


2

The remaining quadratic factor (x − 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros.

So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x + 2)(x + 1)(x – 3 +

)(x – 3 −

). b. g(x) written as a product of linear factors is g(x) = (x + 2)(x + 1)(x – 3 + c. The zeros are –2, –1, 3 −

,3+

)(x – 3 −

).

.

44. h(x) = x4 + 2x3 – 15x2 + 18x – 216 SOLUTION:   a. h(x) has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±27, ±36, ±54, ±72, ±108, and ±216. By using synthetic division, it can be determined that x = 4 is a rational zero.


2

The remaining quadratic factor (x + 9) yields no real zeros and is therefore, irreducible over the reals. So, h(x) 2 written as a product of linear and irreducible quadratic factors is h(x) = (x + 9)(x – 4)(x + 6). 2 b. x + 9 can be written as (x + 3i)(x − 3i). h(x) written as a product of linear factors is h(x) = (x – 3i)(x + 3i)(x – 4)(x + 6). c. The zeros are 3i, –3i, 4, and –6.

45. f (x) = 4x4 – 35x3 + 140x2 – 295x + 156 eSolutions Manual - Powered by Cognero SOLUTION:

a. f (x) has possible rational zeros of ±1, ±2, ±3, ±4, ±12, ±13, ±26, ±39, ±52, ±78,

Page 27

2

2-4

written as a product of linear and irreducible quadratic factors is h(x) = (x + 9)(x – 4)(x + 6). 2 b. x + 9 can be written as (x + 3i)(x − 3i). h(x) written as a product of linear factors is h(x) = (x – 3i)(x + 3i)(x – 4)(x + 6).of Polynomial Functions Zeros c. The zeros are 3i, –3i, 4, and –6.

45. f (x) = 4x4 – 35x3 + 140x2 – 295x + 156 SOLUTION:   a. f (x) has possible rational zeros of ±1, ±2, ±3, ±4, ±12, ±13, ±26, ±39, ±52, ±78, ±156,

By using synthetic division, it can be determined that x = 4 is a

rational zero.

is a rational zero.


2

2

The remaining quadratic factor (4x −16x + 52) can be written 4(x − 4x + 13) and yields no real zeros and is therefore, irreducible over the reals. So, f (x) written as a product of linear and irreducible quadratic factors is 2

or (x – 4x + 13)(4x – 3)(x – 4) b. Use the quadratic formula to find the zeros of x2 −4x + 13.

2

x − 16x + 52 can be written as [x − (2 + 3i)][x − (2 − 3i)]. Thus, f (x) written as a product of linear factors is f (x) = (4x – 3)(x – 4)(x – 2 + 3i)(x – 2 – 3i). c. The zeros are

, 4, 2 – 3i, and 2 + 3i.

46. f (x) = 4x4 – 15x3 + 43x2 + 577x + 615 SOLUTION:   a. f (x) has possible rational zeros of ±1, ±3, ±5, ±15, ±41, ±123, ±205, ±615, ± ±

,±

,±

,±

,±

,±

,±

,±

,±

, and ±

,±

,±

,±

,±

,±

,

. By using synthetic division, it can be

determined that x = −3 is a rational zero.



Page 28

is a rational zero.

2

x − 16x + 52 can be written as [x − (2 + 3i)][x − (2 − 3i)]. Thus, f (x) written as a product of linear factors is f (x) = (4x – 3)(x – 4)(x – 2 + 3i)(x – 2 – 3i).

2-4 c. Zeros of Polynomial The zeros are , 4, 2 – 3iFunctions , and 2 + 3i. 46. f (x) = 4x4 – 15x3 + 43x2 + 577x + 615 SOLUTION:   a. f (x) has possible rational zeros of ±1, ±3, ±5, ±15, ±41, ±123, ±205, ±615, ± ±

,±

,±

,±

,±

,±

,±

,±

,±

, and ±

,±

,±

,±

,±

,±

,

. By using synthetic division, it can be

determined that x = −3 is a rational zero.


2

is a rational zero.

2

The remaining quadratic factor (4x −32x + 164) can be written 4(x − 8x + 41) and yields no real zeros and is therefore, irreducible over the reals. So, f (x) written as a product of linear and irreducible quadratic factors is 2

or (4x + 5)(x + 3)(x – 8x + 41) b. Use the quadratic formula to find the zeros of x2 −8x + 41.

2

x − 8x + 41 can be written as [x − (4 + 5i)][x − (4 − 5i)]. Thus, f (x) written as a product of linear factors is f (x) = (4x + 5)(x + 3)(x – 4 + 5i)(x – 4 – 5i). c. The zeros are

.

47. h(x) = x4 − 2x3 − 17x2 + 4x + 30 SOLUTION:   a. h(x) has possible rational zeros of ±1, ±3, ±5, ±6, ±10, and ±30. By using synthetic division, it can be determined that x = −3 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero. eSolutions Manual - Powered by Cognero

Page 29

2

x − 8x + 41 can be written as [x − (4 + 5i)][x − (4 − 5i)]. Thus, f (x) written as a product of linear factors is f (x) = (4x + 5)(x + 3)(x – 4 + 5i)(x – 4 – 5i).

2-4 c. Zeros of Polynomial Functions. The zeros are 47. h(x) = x4 − 2x3 − 17x2 + 4x + 30 SOLUTION:   a. h(x) has possible rational zeros of ±1, ±3, ±5, ±6, ±10, and ±30. By using synthetic division, it can be determined that x = −3 is a rational zero.


2

The remaining quadratic factor (x − 2) yields no rational zeros and can be written as (x +

)(x −

). So, h(x)

written as a product of linear and irreducible quadratic factors is h(x) = (x + 3)(x − 5)(x +

)(x −

).

b. h(x) written as a product of linear factors is h(x) = (x + 3)(x − 5)(x + c. The zeros are –3, 5,

, and –

)(x −

).

.

48. g(x) = x4 + 31x2 – 180 SOLUTION:   a. g(x) has possible rational zeros of ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, and 2

±180. By using synthetic division, it can be determined that g(x) has no rational zeros. Substitute u = x to factor g (x).

2

2

x − 5 can be written as (x + )(x − ). The remaining quadratic factor (x + 36) yields no real zeros and is therefore, irreducible over the reals. So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x +

)(x −

2

)(x + 36).

b. x2 + 36 can be written as (x + 6i)(x − 6i). g(x) written as a product of linear factors is g(x) = (x + (x + 6i)(x – 6i). c. The zeros are ,– , 6i, and –6i.

)(x −

)

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. h(x) = 2x5 + x4 – 7x3 + 21x2 – 225x + 108; 3i

SOLUTION:   Use synthetic substitution to that 3i is a zero of h(x).


Because x = 3i is a zero of h, x = −3i is also a zero of h. Divide the depressed polynomial by −3i.

Page 30

(x +

2-4

2

)(x −

)(x + 36).

b. x2 + 36 can be written as (x + 6i)(x − 6i). g(x) written as a product of linear factors is g(x) = (x + (x + 6i)(x – 6i). Zeros of Polynomial Functions c. The zeros are ,– , 6i, and –6i.

)(x −

)

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. h(x) = 2x5 + x4 – 7x3 + 21x2 – 225x + 108; 3i

SOLUTION:   Use synthetic substitution to that 3i is a zero of h(x).

Because x = 3i is a zero of h, x = −3i is also a zero of h. Divide the depressed polynomial by −3i.

3

2

Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 3i)(x − 3i)(2x + x − 3

2

25x + 12). 2x + x − 25x + 12 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±12,

. By using

synthetic division, it can be determined that x = 3 is a rational zero.

2

The remaining depressed polynomial 2x +7x − 4 can be written as (x + 4)(2x − 1). The zeros of the depressed polynomial are −4 and

. Therefore, the zeros of h are 3, –4,

, 3i, and –3i. The linear factorization of h is h(x) =

(x – 3)(x + 4)(2x – 1)(x + 3i)(x – 3i).

50. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x + 1500; –5i SOLUTION:   Use synthetic substitution to that −5i is a zero of h(x).

Because x = −5i is a zero of h, x = 5i is also a zero of h. Divide the depressed polynomial by 5i.

3

2

Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 5i)(x − 5i)(3x − 5x − 3 2 88x + 60). 3x − 5x − 88x + 60 has possible rational zeros of ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± ,±

,±

,± ,±

, and ±

. By using synthetic division, it can be determined that x = −5 is a rational zero.

2 eSolutions - Powered by Cognero TheManual remaining depressed polynomial 3x

polynomial are 6 and

Page 31 − 20x + 12 can be written as (x − 6)(3x − 2). The zeros of the depressed

. Therefore, the zeros of h are –5, 6,


2

The remaining depressed polynomial 2x +7x − 4 can be written as (x + 4)(2x − 1). The zeros of the depressed polynomial are −4 and

. Therefore, the zeros of h are 3, –4,


2-4 (x Zeros Polynomial Functions – 3)(xof + 4)(2x – 1)(x + 3i)(x – 3i). 50. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x + 1500; –5i SOLUTION:   Use synthetic substitution to that −5i is a zero of h(x).

Because x = −5i is a zero of h, x = 5i is also a zero of h. Divide the depressed polynomial by 5i.

3

2

Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 5i)(x − 5i)(3x − 5x − 3 2 88x + 60). 3x − 5x − 88x + 60 has possible rational zeros of ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± ,±

,±

,± ,±

, and ±

. By using synthetic division, it can be determined that x = −5 is a rational zero.

2

The remaining depressed polynomial 3x − 20x + 12 can be written as (x − 6)(3x − 2). The zeros of the depressed polynomial are 6 and



(x + 5)(x – 6)(3x – 2)(x + 5i)(x – 5i).

51. g(x) = x5 – 2x4 – 13x3 + 28x2 + 46x – 60; 3 – i SOLUTION:   Use synthetic substitution to that 3 − i is a zero of g(x).

Because x = 3 − i is a zero of g, x = 3 + i is also a zero of g. Divide the depressed polynomial by 3 + i.

3

Using these two zeros and the depressed polynomial from the last division, write h(x) = [x − (3 − i)][x − (3 + i)](x + 2 3 2 4x + x − 6). x + 4x + x − 6 has possible rational zeros of ±1, ±2, ±3, and ±6. By using synthetic division, it can be determined that x = 1 is a rational zero.

2

The remaining depressed polynomial x + 5x + 6 can be written as (x + 3)(x + 2). The zeros of the depressed polynomial are −3 and −2. Therefore, the zeros of g are –3, –2, 1, 3 + i, and 3 – i. The linear factorization of g is g (x) = (x + 3)(x + 2)(x – 1)(x – 3 + i)(x – 3 – i). eSolutions Manual - Powered by Cognero

52. g(x) = 4x5 – 57x4 + 287x3 – 547x2 + 83x + 510; 4 + i SOLUTION:

Page 32

2

The remaining depressed polynomial 3x − 20x + 12 can be written as (x − 6)(3x − 2). The zeros of the depressed polynomial are 6 and



2-4 Zeros of Polynomial Functions (x + 5)(x – 6)(3x – 2)(x + 5i)(x – 5i).

51. g(x) = x5 – 2x4 – 13x3 + 28x2 + 46x – 60; 3 – i SOLUTION:   Use synthetic substitution to that 3 − i is a zero of g(x).

Because x = 3 − i is a zero of g, x = 3 + i is also a zero of g. Divide the depressed polynomial by 3 + i.

3

Using these two zeros and the depressed polynomial from the last division, write h(x) = [x − (3 − i)][x − (3 + i)](x + 2 3 2 4x + x − 6). x + 4x + x − 6 has possible rational zeros of ±1, ±2, ±3, and ±6. By using synthetic division, it can be determined that x = 1 is a rational zero.

2

The remaining depressed polynomial x + 5x + 6 can be written as (x + 3)(x + 2). The zeros of the depressed polynomial are −3 and −2. Therefore, the zeros of g are –3, –2, 1, 3 + i, and 3 – i. The linear factorization of g is g (x) = (x + 3)(x + 2)(x – 1)(x – 3 + i)(x – 3 – i).

52. g(x) = 4x5 – 57x4 + 287x3 – 547x2 + 83x + 510; 4 + i SOLUTION:   Use synthetic substitution to that 4 + i is a zero of g(x).

Because x = 4 + i is a zero of g, x = 4 − i is also a zero of g. Divide the depressed polynomial by 4 − i.

Using these two zeros and the depressed polynomial from the last division, write h(x) = [x − (4 + i)][x − (4 − i)](4x 2 3 2 − 25x + 19x + 30). 4x − 25x + 19x + 30 has possible rational zeros of ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30,

3


2

The remaining depressed polynomial 4x − 17x − 15 can be written as (x − 5)(4x + 3). The zeros of the depressed polynomial are 5 and

. Therefore, the zeros of g are

(x – 5)(4x + 3)(x – 2)(x – 4 – i)(x – 4 + i).


53. f (x) = x5 – 3x4 – 4x3 + 12x2 – 32x + 96; –2i

. The linear factorization of g is g(x) = Page 33

2

2-4

The remaining depressed polynomial x + 5x + 6 can be written as (x + 3)(x + 2). The zeros of the depressed polynomial the zeros of g are –3, –2, 1, 3 + i, and 3 – i. The linear factorization of g is g −3 and −2. Therefore, Zeros ofare Polynomial Functions (x) = (x + 3)(x + 2)(x – 1)(x – 3 + i)(x – 3 – i).

52. g(x) = 4x5 – 57x4 + 287x3 – 547x2 + 83x + 510; 4 + i SOLUTION:   Use synthetic substitution to that 4 + i is a zero of g(x).


Using these two zeros and the depressed polynomial from the last division, write h(x) = [x − (4 + i)][x − (4 − i)](4x 2 3 2 − 25x + 19x + 30). 4x − 25x + 19x + 30 has possible rational zeros of ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30,

3


2



. The linear factorization of g is g(x) =

(x – 5)(4x + 3)(x – 2)(x – 4 – i)(x – 4 + i).

53. f (x) = x5 – 3x4 – 4x3 + 12x2 – 32x + 96; –2i SOLUTION:   Use synthetic substitution to that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

3

2

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x − 3x − 8x 3 2 + 24). x − 3x − 8x + 24 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24. By using synthetic division, it can be determined that x = 3 is a rational zero.

2

The remaining depressed polynomial x − 8 can be written as (x − zeros of the depressed polynomial are 2 eSolutions Manual - Powered by Cognero

and −2

The linear factorization of f is f (x) = (x – 3)(x + 2

54. g(x) = x4 – 10x3 + 35x2 – 46x + 10; 3 + i

)(x +

) or (x − 2

. Therefore, the zeros of f are 3, 2 )(x − 2

)(x + 2i)(x – 2i).

)(x + 2 , –2

). The , 2i, and –2i. Page 34

2



. The linear factorization of g is g(x) =

2-4 Zeros of Polynomial Functions

(x – 5)(4x + 3)(x – 2)(x – 4 – i)(x – 4 + i).

53. f (x) = x5 – 3x4 – 4x3 + 12x2 – 32x + 96; –2i SOLUTION:   Use synthetic substitution to that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

3

2

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x − 3x − 8x 3 2 + 24). x − 3x − 8x + 24 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24. By using synthetic division, it can be determined that x = 3 is a rational zero.

2

The remaining depressed polynomial x − 8 can be written as (x − zeros of the depressed polynomial are 2

and −2


)(x +

) or (x − 2


)(x + 2 , –2

). The , 2i, and –2i.

)(x + 2i)(x – 2i).

54. g(x) = x4 – 10x3 + 35x2 – 46x + 10; 3 + i SOLUTION:   Use synthetic substitution to that 3 + i is a zero of g(x).


2

Using these two zeros and the depressed polynomial from the last division, write g(x) = [x − (3 + i)][x − (3 − i)](x − 2

4x + 1). x − 4x + 1 yields no rational zeros. Use the quadratic formula to find the zeros.

eSolutions TheManual zeros -ofPowered g are 2by+Cognero, 2

)(x – 3 – i)(x – 3 + i).

−

, 3 + i, and 3 − i. The linear factorization of g is g(x) = (x – 2 −

)(x – 2Page + 35

2

The remaining depressed polynomial x − 8 can be written as (x − zeros of the depressed polynomial are 2


and −2


)(x +

) or (x − 2


)(x + 2 , –2

). The , 2i, and –2i.

)(x + 2i)(x – 2i).

54. g(x) = x4 – 10x3 + 35x2 – 46x + 10; 3 + i SOLUTION:   Use synthetic substitution to that 3 + i is a zero of g(x).


2

Using these two zeros and the depressed polynomial from the last division, write g(x) = [x − (3 + i)][x − (3 − i)](x − 2

4x + 1). x − 4x + 1 yields no rational zeros. Use the quadratic formula to find the zeros.

The zeros of g are 2 +

,2 −

, 3 + i, and 3 − i. The linear factorization of g is g(x) = (x – 2 −

)(x – 2 +

)(x – 3 – i)(x – 3 + i).

55. ARCHITECTURE  An architect is constructing a scale model of a building that is in the shape of a pyramid. a. If the height of the scale model is 9 inches less than its length and its base is a square, write a polynomial function that describes the volume of the model in of its length. b. If the volume of the model is 6300 cubic inches, write an equation describing the situation. c. What are the dimensions of the scale model?

SOLUTION:   a. The volume of a pyramid is V =

Bh, where B is the area of the base. Let l represent the length of one of the 2

sides of the square base. The area of the base B is B = l . The height h is h = and h into the formula for the volume of a pyramid.


−9. Substitute these values for B

A polynomial function that describes the volume of the model in of its length is V(l) = b. Substitute V(l) = 6300 into the equation found in part a.

3

l –3

2

.

Page 36

2-4 Zeros of Polynomial Functions A polynomial function that describes the volume of the model in of its length is V(l) =

3

l –3

2

.

b. Substitute V(l) = 6300 into the equation found in part a. 6300 =

3

l –3

2

c. Solve the equation found in part b for

.

This polynomial has 1 sign variation, so it has 1 positive real zero. The graph suggests that

= 30 is a real zero.

Use synthetic division to test this possibility.

must be a positive value. Because V( ) only has 1 positive real zero, no other tests are necessary. Thus, the base is 30 inches by 30 inches and the height of the model is 21 inches.

56. CONSTRUCTION  The height of a tunnel that is under construction is 1 foot more than half its width and its length is 32 feet more than 324 times its width. If the volume of the tunnel is 62,231,040 cubic feet and it is a rectangular prism, find the length, width, and height.

SOLUTION:   The volume of a rectangular prism is V = lwh. The height h is h = 1 +

and the width is w = 32 + 324w. Substitute

these values into the equation for the volume of the tunnel.

Substitute V(w) = 62,231,040 in the equation and solve for w.

3 Cognero 2 eSolutions - Powered TheManual polynomial 81wby + 170w

Page 37 + 16w − 31,115,520 has 1 sign variation, so it has 1 positive real zero. Using the zero function from the CALC menu suggests that w = 72 is a real zero.

must be a positive value. Because V( 2-4 Zeros of Polynomial Functions) only has 1 positive real zero, no other tests are necessary. Thus, the base is 30 inches by 30 inches and the height of the model is 21 inches.

56. CONSTRUCTION  The height of a tunnel that is under construction is 1 foot more than half its width and its length is 32 feet more than 324 times its width. If the volume of the tunnel is 62,231,040 cubic feet and it is a rectangular prism, find the length, width, and height.

SOLUTION:   The volume of a rectangular prism is V = lwh. The height h is h = 1 +

and the width is w = 32 + 324w. Substitute

these values into the equation for the volume of the tunnel.

Substitute V(w) = 62,231,040 in the equation and solve for w.

3

2

The polynomial 81w + 170w + 16w − 31,115,520 has 1 sign variation, so it has 1 positive real zero. Using the zero function from the CALC menu suggests that w = 72 is a real zero.

Use synthetic division to test this possibility.

w must be a positive value. Because V(w) only has 1 positive real zero, no other tests are necessary. Thus, the width is 72 feet, the height is 1 +

or 37 feet, and the length is 32 + 324(72) or 23,360 feet.

Write a polynomial function of least degree with integer coefficients that has the given number as a zero.

57. SOLUTION:   Sample answer: Using the Linear Factorization Theorem and the zero Let a = 1. Then write the function in standard form. Since is a factor, when eSolutions Manual - Powered by Cognero


. Page 38

w must be a positive value. Because V(w) only has 1 positive real zero, no other tests are necessary. Thus, the width

2-4 is Zeros 72 feet,ofthePolynomial height is 1 + Functions or 37 feet, and the length is 32 + 324(72) or 23,360 feet. Write a polynomial function of least degree with integer coefficients that has the given number as a zero.

57. SOLUTION:   Sample answer: Using the Linear Factorization Theorem and the zero Let a = 1. Then write the function in standard form. Since is a factor, when


.

3

as a zero is f (x) = x − 6.

A polynomial function of least degree with integer coefficients that has

58. SOLUTION:   Sample answer: Using the Linear Factorization Theorem and the zero Let a = 1. Then write the function in standard form. . Since is a factor,

when


.

3




when

A polynomial function of least degree with integer coefficients that has eSolutions Manual - Powered by Cognero

60.


.

3

as a zero is f (x) = x + 2. Page 39


3




when


.

3

as a zero is f (x) = x + 2.



when


.


3


Use each graph to write g as the product of linear factors. Then list all of its zeros. 61. g(x) = 3x4 – 15x3 + 87x2 – 375x + 300

SOLUTION:   The graph suggests 1 and 4 are zeros of g(x). Use synthetic substitution to test this possibility.

Use synthetic division on the depressed polynomial to test 4. eSolutions Manual - Powered by Cognero

Page 40

2

2



3


Use each graph to write g as the product of linear factors. Then list all of its zeros. 61. g(x) = 3x4 – 15x3 + 87x2 – 375x + 300

SOLUTION:   The graph suggests 1 and 4 are zeros of g(x). Use synthetic substitution to test this possibility.

Use synthetic division on the depressed polynomial to test 4.

2

2

The remaining quadratic factor (3x + 75) can be written as 3(x + 25) or 3(x + 5i)(x − 5i). So, g written as the product of linear factors is g(x) = 3(x – 4)(x – 1)(x + 5i)(x – 5i). The zeros of g are 4, 1, 5i.

62. g(x) = 2x5 + 2x4 + 28x3 + 32x2 – 64x

SOLUTION:   The graph suggests −2 and 1 are zeros of g(x). Use synthetic substitution to test this possibility.

Use synthetic division on the depressed polynomial to test −2.

3

2

The remaining factor (2x + 32x) can be written as 2x(x + 16) or 2x(x + 4i)(x − 4i). So, g written as the product of linear factors is g(x) = 2x(x – 1)(x + 2)(x + 4 i)(x – 4 i). The zeros of g are 0, 1, –2, 4i. Determine all rational zeros of the function. 63. h(x) = 6x3 − 6x2 + 12

SOLUTION:   eSolutions - Powered by Cognero TheManual leading coefficient is 6 and

±2, ±3, ±4, ±6, ±12, ±

,±

,

the constant term is 12. The possible rational zeros are ,±

,±

, and ±

.

Page 41 or ±1,

3

2

+ 16) or 2x(x + 4i)(x − 4i). So, g written as the product of linear factors is g(x) = 2x(x – 1)(x + 2)(x + 4 i)(x – 4 i). The zeros of g are 0, 1, –2, 4i.

remaining factor (2x + 32x) can be written as 2x(x 2-4 The Zeros of Polynomial Functions Determine all rational zeros of the function. 63. h(x) = 6x3 − 6x2 + 12

SOLUTION:   The leading coefficient is 6 and the constant term is 12. The possible rational zeros are ±2, ±3, ±4, ±6, ±12, ±

,±

,±

,

,±

, and ±

or ±1,

.


2

2

2

The remaining quadratic factor (6x − 12x + 12) can be written as 6(x − 2x + 2). x − 2x + 2 yields no rational zeros. Thus, the only rational zero of h is −1.

64. SOLUTION:   To find the zeros of f (y), set f (y) = 0 and multiply the equation by 4.

By the Linear Factorization Theorem, this polynomial will have the same zeros as f (y). Because the leading coefficient is 1, the possible rational zeros of the new function are the integer factors of the constant term −32. Therefore, the possible rational zeros are ±1, ±2, ±4, ±8, ±16, and ±32. By using synthetic division, it can be determined that x = 2 is a rational zero.


2

The remaining quadratic factor (x + 4) yields no real zeros. Thus, the rational zeros of f are −4 and 2.

65. w(z) = z 4 − 10z 3 + 30z 2 − 10z + 29 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros of g are ±1, ±2, ±4, and ±8. However, the graph suggests that w(z) has no rational zeros. eSolutions Manual - Powered by Cognero

Page 42

2-4 Zeros of Polynomial Functions 2

The remaining quadratic factor (x + 4) yields no real zeros. Thus, the rational zeros of f are −4 and 2.

65. w(z) = z 4 − 10z 3 + 30z 2 − 10z + 29 SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros of g are ±1, ±2, ±4, and ±8. However, the graph suggests that w(z) has no rational zeros.

Testing each of the possibilities using synthetic division that w(z) has no rational zeros.

66. SOLUTION:   To find the zeros of b(a), set b(a) = 0 and multiply the equation by 6.

By the Linear Factorization Theorem, this polynomial will have the same zeros as b(a). The leading coefficient is 6 or ±1, ±

and the constant term is 1. The possible rational zeros are

, ± , and

.

By using synthetic division, it can be determined that x = 1 is a rational zero.


By using synthetic division on the new depressed polynomial, it can be determined that

2

is a rational zero.

is a rational zero.

2

The remaining quadratic factor (6x + 6) can be written as 6(x + 1) and yields no real zeros. Thus, the rational eSolutions Manual - Powered by Cognero

zeros of b are

Page 43

67. ENGINEERING  A steel beam is ed by two pilings 200 feet apart. If a weight is placed x feet from the

2-4 Zeros of Polynomial Functions Testing each of the possibilities using synthetic division that w(z) has no rational zeros.

66. SOLUTION:   To find the zeros of b(a), set b(a) = 0 and multiply the equation by 6.

By the Linear Factorization Theorem, this polynomial will have the same zeros as b(a). The leading coefficient is 6 or ±1, ±

and the constant term is 1. The possible rational zeros are

, ± , and

.




2

is a rational zero.

is a rational zero.

2

The remaining quadratic factor (6x + 6) can be written as 6(x + 1) and yields no real zeros. Thus, the rational zeros of b are

67. ENGINEERING  A steel beam is ed by two pilings 200 feet apart. If a weight is placed x feet from the 2

piling on the left, a vertical deflection represented by d = 0.0000008x (200 – x) occurs. How far is the weight from the piling if the vertical deflection is 0.8 feet?

SOLUTION:   2

Solve d = 0.0000008x (200 – x) for d = 0.8. eSolutions Manual - Powered by Cognero

Page 44

2

2

The remaining quadratic factor (6x + 6) can be written as 6(x + 1) and yields no real zeros. Thus, the rational

2-4 zeros Zeros Polynomial Functions of bofare 67. ENGINEERING  A steel beam is ed by two pilings 200 feet apart. If a weight is placed x feet from the 2

piling on the left, a vertical deflection represented by d = 0.0000008x (200 – x) occurs. How far is the weight from the piling if the vertical deflection is 0.8 feet?

SOLUTION:   2

Solve d = 0.0000008x (200 – x) for d = 0.8.

2

Use a graphing calculator to graph y = 0.0000008x (200 − x) − 0.8 and solve for y = 0. Using the CALC menu, find the zeros of the function.

The zeros of the function occur at x = 100 and x = 161.8. A vertical deflection of 0.8 feet will occur when a weight is placed 100 feet or about 161.8 feet from the left piling. Write each polynomial as the product of linear and irreducible quadratic factors. 68. x3 – 3

SOLUTION:   3

3

2

2

3

Use the difference of two cubes, a − b = (a − b)(a + ab + b ), to factor x − 3.

69. x3 + 16 SOLUTION:   3

3

2

2

3

3

2

2

3

Use the sum of two cubes, a + b = (a + b)(a − ab + b ), to factor x + 16.

70. 8x3 + 9 SOLUTION:   3

Use the sum of two cubes, a + b = (a + b)(a − ab + b ), to factor 8x + 9. eSolutions Manual - Powered by Cognero

Page 45

3

3

2

2

3

Use the sum of two cubes, a + b = (a + b)(a − ab + b ), to factor x + 16.

2-4 Zeros of Polynomial Functions 70. 8x3 + 9 SOLUTION:   3

3

2

2

3

3

2

2

3

Use the sum of two cubes, a + b = (a + b)(a − ab + b ), to factor 8x + 9.

71. 27x6 + 4 SOLUTION:   6

Use the sum of two cubes, a + b = (a + b)(a − ab + b ), to factor 27x + 4.

72. MULTIPLE REPRESENTATIONS In this problem, you will explore even- and odd-degree polynomial functions. a. ANALYTICAL  Identify the degree and number of zeros of each polynomial function. i. f (x) = x3 – x2 + 9x – 9 5 4 ii. g(x) = 2x + x – 32x – 16 iii. h(x) = 5x3 + 2x2 – 13x + 6 4

2

iv. f (x) = x + 25x + 144 v. h(x) = 3x6 + 5x5 + 46x4 +80x3 – 32x2 4 3 2 vi. g(x) = 4x – 11x + 10x – 11x + 6 b. NUMERICAL  Find the zeros of each function. c. VERBAL  Does an odd-degree function have to have a minimum number of real zeros? Explain.

SOLUTION:   a. i. The degree is 3. Thus, f (x) will have 3 zeros. ii. The degree is 5, Thus, g(x) will have 5 zeros. iii. The degree is 3, Thus, h(x) will have 3 zeros. iv. The degree is 4, Thus, f (x) will have 4 zeros. v. The degree is 6, Thus, h(x) will have 6 zeros. vi. The degree is 4, Thus, g(x) will have 4 zeros. b. i. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −9. Therefore, the possible rational zeros of f are ±1, ±3, and ±9. By using synthetic division, it can be determined that x = 1 is a rational zero.

2 The remaining quadratic factor (x + 9) can be written as (x + 3i)(x − 3i). Thus, the zeros of f are 3i, –3i, and 1.

ii. The leading coefficient is 2 and the constant term is −16. The possible rational zeros are ±2, ±6, ±8, ±16, and

or ±1,

.

By using synthetic division, it can be determined that x = 2 is a rational zero. eSolutions Manual - Powered by Cognero

Page 46

2

The remaining quadratic factor (x + 9) can be written as (x + 3i)(x − 3i). Thus, the zeros of f are 3i, –3i, and 1. ii. The leading coefficient is 2 and the constant term is −16. The possible rational zeros are

or ±1,

2-4 Zeros of Polynomial Functions ±2, ±6, ±8, ±16, and

.



is a rational zero.


2

The remaining quadratic factor (2x + 8) can be written as 2(x + 2i)(x − 2i). Thus, the zeros of g are

iii. The leading coefficient is 5 and the constant term is 6. The possible rational zeros are ±3, ±6,

or ±1, ±2,

.


2

The remaining quadratic factor (5x + 7x − 6) can be written as (x + 2)(5x − 3). Thus, the zeros of h are –2,

, and

1. iv. Substitute u = x2 and factor f (x).

2

2

The factor (x + 9) can be written as (x + 3i)(x − 3i) and the factor (x + 16) can be written as (x + 4i)(x − 4i). Thus, the zeros of f are –3i, 3i, –4i, and 4i. v. h(x) can be written as h(x) = x2(3x4 + 5x3 + 46x2 + 80x − 32). Find the zeros of the polynomial 3x4 + 5x3 + 46x2 + 80x − 32. The leading coefficient is 3 and the constant term is −32. The possible rational zeros are or ±1, ±2, ±4, ±8, ±16, ±32,

.




Page 47

is a rational zero.

v. h(x) can be written as h(x) = x2(3x4 + 5x3 + 46x2 + 80x − 32). Find the zeros of the polynomial 3x4 + 5x3 + 46x2 + 80x − 32. The leading coefficient is 3 and the constant term is −32. The possible rational zeros are or ±1,Functions ±2, ±4, ±8, ±16, ±32, 2-4 Zeros of Polynomial

.



is a rational zero.

2

The remaining quadratic factor (3x + 48) can be written as 3(x + 4i)(x − 4i). Thus, the zeros of h are 4i, –4i, −2, 0 (multiplicity: 2), and

.

vi. The leading coefficient is 4 and the constant term is 6. The possible rational zeros are ±3, ±6,

or ±1, ±2,

.



is a rational zero.

2

The remaining quadratic factor (4x + 4) can be written as 4(x + i)(x − i). Thus, the zeros of g are i, – i,

, and 2.

c. An odd-degree polynomial function always has an odd number of zeros, and a polynomial function with real coefficients has imaginary zeros that occur in conjugate pairs. Therefore, an odd function with real coefficients will always have at least one real zero.

73. ERROR ANALYSIS Angie and Julius are using the Rational Zeros Theorem to find all the possible rational zeros 2

3

of f (x) = 7x +2x – 5x – 3. Angie thinks the possible zeros are

, ±1, ±3. Julius thinks they are

, ±1,

±3. Is either of them correct? Explain your reasoning.

SOLUTION:   Sample answer: Angie divided by the factors of 7, which is not the leading coefficient. Julius divided by the factors of 2, which is the leading coefficient. Thus, Julius is correct.

74. REASONING Explain why g(x) = x9 − x8 + x5 + x3 − x2 + 2 must have a root between x = −1 and x = 0. SOLUTION:   Sample answer: Evaluate g(x) for x = −1 and x = 0. eSolutions Manual - Powered by Cognero

Page 48

±3. Is either of them correct? Explain your reasoning.

SOLUTION:   Angie divided by the factors of 7, which is not the leading coefficient. Julius divided by the factors 2-4 Sample Zerosanswer: of Polynomial Functions of 2, which is the leading coefficient. Thus, Julius is correct.

74. REASONING Explain why g(x) = x9 − x8 + x5 + x3 − x2 + 2 must have a root between x = −1 and x = 0. SOLUTION:   Sample answer: Evaluate g(x) for x = −1 and x = 0.

An input of –1 produces a negative output, whereas an input of 0 results in a positive output.

75. CHALLENGE Use f (x) = x2 + x − 6, f (x) = x3 + 8x2 + 19x + 12, and f (x) = x4 − 2x3 − 21x2 + 22x + 40 to make a conjecture about the relationship between the graphs and zeros of f (x) and the graphs and zeros of each of the following. a. −f (x) b. f (−x)

SOLUTION:   a. Graph each f (x) and its corresponding −f (x).

The graphs of −f (x) are the graphs of f (x) reflected in the x-axis. The zeros are the same. b. Graph each f (x) and its corresponding f (−x).


Page 49

2-4 The Zeros of Polynomial Functions graphs of −f (x) are the graphs of f (x) reflected in the x-axis. The zeros are the same. b. Graph each f (x) and its corresponding f (−x).

The graphs of f (−x) are the graphs of f (x) reflected in the y-axis. The zeros are opposites.

76. OPEN ENDED Write a function of 4th degree with an imaginary zero and an irrational zero. SOLUTION:   Sample answer: A factor with an imaginary zero is (x + i). Since (x + i) is a factor, (x − i) must also be a factor. A ). Since the function is to have a degree of 4, let (x −

factor with an irrational root is (x + f(x) = (x + i) (x − i)(x + 4

)(x −

) be a root.

)

2

       = x − 2x − 3

77. REASONING Determine whether the statement is true or false . If false, provide a counterexample. A third-degree polynomial with real coefficients has at least one nonreal zero.

SOLUTION:   3

2

The statement is false. Sample answer: The third-degree polynomial x + x − 17x + 15 has three real zeros and no nonreal zeros. CHALLENGE  Find the zeros of each function if h(x) has zeros at x 1, x 2, and x 3.

78. c(x) = 7h(x) SOLUTION:   Let h(x) = (x − x1)(x − x2)(x − x3). Multiply h(x) by 7. h(x) = 7(x − x1)(x − x2)(x − x3) Multiplying h(x) by 7 will result in a vertical dilation that will expand the graph, but will not change the zeros. Thus, the zeros of c(x) are x1, x2, and x3.

79.

=

k(x)Manual h(3x) eSolutions - Powered by Cognero

SOLUTION:   Let h(x) = (x − x )(x − x )(x − x ). Substitute 3x for x.

Page 50

1

2

3

h(x) = 7(x − x1)(x − x2)(x − x3)

2-4

Multiplying h(x) by 7 will result in a vertical dilation that will expand the graph, but will not change the zeros. Thus, the zerosof of Polynomial c(x) are x1, x2, and x3. Zeros Functions

79. k(x) = h(3x) SOLUTION:   Let h(x) = (x − x1)(x − x2)(x − x3). Substitute 3x for x. h(x) = (3x − x1)(3x − x2)(3x − x3) Find zero for each factor.

Multiplying x by 3 will result in a horizontal dilation that will compress the graph and change the zeros. Thus, the zeros of k(x) are

.

80. g(x) = h(x – 2) SOLUTION:   Let h(x) = (x − x1)(x − x2)(x − x3). Substitute x − 2 for x. h(x) = [(x − 2) − x1][(x − 2) − x2][(x − 2) − x3] Find zero for each factor.

Subtracting 2 from x will result in a horizontal shift that will shift the zeros. Thus, the zeros of g(x) are x1  + 2, x2 + 2, and x3 + 2.

81. f (x) = h(–x) SOLUTION:   Let h(x) = (x − x1)(x − x2)(x − x3). Substitute −x for x. h(x) = (−x − x1)(−x − x2)(−x − x3) Find zero for each factor.

Multiplying x by −1 will reflect the graph of h(x) in the y-axis. Thus, the zeros of f (x) are –x1, –x2, and –x3.

82. REASONING If x − c is a factor of f (x) = a 1x5 − a 2x4 + …, what value must c be greater than or equal to in order to be an upper bound for the zeros of f (x)? Assume a ≠ 0. Explain your reasoning.

SOLUTION:   Sample answer: If c is an upper bound for the zeros of f (x), then every value in the last line of the synthetic division is nonnegative. eSolutions Manual - Powered by Cognero

Page 51

2-4 Multiplying Zeros ofxPolynomial Functions by −1 will reflect the graph of h(x) in the y-axis. Thus, the zeros of f (x) are –x

1, –x2, and –x3.

82. REASONING If x − c is a factor of f (x) = a 1x5 − a 2x4 + …, what value must c be greater than or equal to in order to be an upper bound for the zeros of f (x)? Assume a ≠ 0. Explain your reasoning.

SOLUTION:   Sample answer: If c is an upper bound for the zeros of f (x), then every value in the last line of the synthetic division is nonnegative.

Thus, ca1 − a 2 ≥ 0. Solve for c.

c must be greater than or equal to

. If c is less than

, then the second term of the depressed polynomial will

be negative and the upper bound test will fail.

83. Writing in Math Explain why a polynomial with real coefficients and one imaginary zero must have at least two imaginary zeros.

SOLUTION:   Sample answer: If a polynomial has an imaginary zero, then its complex conjugate is also a zero of the polynomial. Therefore, any polynomial that has one imaginary zero has at least two imaginary zeros. Divide using synthetic division.

84. (x3 – 9x2 + 27x – 28) ÷ (x – 3) SOLUTION:   Because x − 3, c = 3. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is

.

85. (x4 + x3 – 1) ÷ (x – 2) SOLUTION:   Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

3

2

The quotient is x + 3x + 6x + 12 + eSolutions Manual - Powered by Cognero

86. (3x4 – 2x3 + 5x2 – 4x – 2) ÷ (x + 1) SOLUTION:

. Page 52

quotient is . 2-4 The Zeros of Polynomial Functions

85. (x4 + x3 – 1) ÷ (x – 2) SOLUTION:   Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

3

2

The quotient is x + 3x + 6x + 12 +

.

86. (3x4 – 2x3 + 5x2 – 4x – 2) ÷ (x + 1) SOLUTION:   Because x + 1, c = −1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

3

2

The quotient is 3x – 5x + 10x – 14 +

.

87. (2x3 – 2x – 3) ÷ (x – 1) SOLUTION:   Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is

.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

88. f (x) = –4x7 + 3x4 + 6 SOLUTION:   The degree is 7, and the leading coefficient is −4. Because the degree is odd and the leading coefficient is negative,

89. f (x) = 4x6 + 2x5 + 7x2 SOLUTION:   The degree is 6, and the leading coefficient is 4. Because the degree is even and the leading coefficient is positive,

90. g(x) = 3x4 + 5x5 – 11 SOLUTION:   eSolutions Manual - Powered by Cognero

Page 53 The degree is 5, and the leading coefficient is 5. Because the degree is odd and the leading coefficient is positive, .

89. f (x) = 4x + 2x + 7x SOLUTION:   The degree is 6, and the leading coefficient is 4. Because the degree is even and the leading coefficient is positive,

2-4 Zeros of Polynomial Functions 90. g(x) = 3x4 + 5x5 – 11 SOLUTION:

The degree is 5, and the leading coefficient is 5. Because the degree is odd and the leading coefficient is positive, . Estimate to the nearest 0.5 unit and classify the extrema for the graph of each function. the answers numerically.

91. SOLUTION:   It appears that f (x) has a relative minimum of –3 at x = 0 and a relative maximum of 3 at x = –2. Create a table. Choose x-values on either side of the estimated values. x 0 0.5 10 −10 −2.5 −2 −1.5 −0.5 f (x) 2.6 3 1.9 1287 −693 −1.9 −3 −2.6 The table s these estimates.

92. SOLUTION:   It appears that f (x) has a relative maximum of 8 at x = 1, and a relative minimum of –16 at x = 3 and x = –1. Create a table. Choose x-values on either side of the estimated values. x 0.5 1 1.5 2.5 3 3.5 10 −10 −1.5 −1 −0.5 f(x) 13,440 −14.4 −15 −8.4 6.6 9 6.6 5760 −8.4 −15 −14.4 The table s these estimates.

93. SOLUTION:   maximum of 0 at x = 0 and x = –4, a relative minimum of –80 at x = –2 and aPage 54 relative minimum of –150 at x = 2. Create a table. Choose x-values on either side of the estimated values.

It appears f (x) by has a relative eSolutions Manual -that Powered Cognero

2-4

It appears that f (x) has a relative maximum of 8 at x = 1, and a relative minimum of –16 at x = 3 and x = –1. Create a table. Choose x-values on either side of the estimated values. x 0.5 1 1.5 2.5 3 3.5 10 −10 −1.5 −1 −0.5 f(x) 13,440 6.6 9 6.6 5760 −14.4 −15Functions −8.4 −8.4 −15 −14.4 Zeros of Polynomial The table s these estimates.

93. SOLUTION:   It appears that f (x) has a relative maximum of 0 at x = 0 and x = –4, a relative minimum of –80 at x = –2 and a relative minimum of –150 at x = 2. Create a table. Choose x-values on either side of the estimated values. x 0 0.5 1.5 2 2.5 −4.5 −4 −3.5 −2.5 −2 −1.5 −0.5 f (x) 0 0 −38 −19.9 −77.3 −80 −63.3 −10.7 −12.7 −102.1 −144 −132 The table s these estimates.

94. FINANCE Investors choose different stocks to comprise a balanced portfolio. The matrices show the prices of one share of each of several stocks on the first business day of July, August, and September.

a. Mrs. Rivera owns 42 shares of stock A, 59 shares of stock B, 21 shares of stock C, and 18 shares of stock D. Write a row matrix to represent Mrs. Rivera’s portfolio. b. Use matrix multiplication to find the total value of Mrs. Rivera’s portfolio for each month to the nearest cent.

SOLUTION:   a. The row will have four columns, one for each stock. Let the first column represent stock A, the second column represent stock B, the third column represent stock C, and the fourth column represent stock D. [42   59   21   18] b. For each month, the prices of each stock can be displayed in matrices.

To find the total value of Mrs. Rivera’s portfolio for each month, multiply the the matrices.


matrix found in part a by each of

Page 55

2-4

relative minimum of –150 at x = 2. Create a table. Choose x-values on either side of the estimated values. x 0 −4.5 −4 −3.5 −2.5 −2 −1.5 −0.5 f (x) 0 0 −19.9 −77.3 −80 −63.3 −10.7 Zeros of−38 Polynomial Functions The table s these estimates.

0.5 −12.7

1.5 −102.1

2 −144

2.5 −132

94. FINANCE Investors choose different stocks to comprise a balanced portfolio. The matrices show the prices of one share of each of several stocks on the first business day of July, August, and September.

a. Mrs. Rivera owns 42 shares of stock A, 59 shares of stock B, 21 shares of stock C, and 18 shares of stock D. Write a row matrix to represent Mrs. Rivera’s portfolio. b. Use matrix multiplication to find the total value of Mrs. Rivera’s portfolio for each month to the nearest cent.

SOLUTION:   a. The row will have four columns, one for each stock. Let the first column represent stock A, the second column represent stock B, the third column represent stock C, and the fourth column represent stock D. [42   59   21   18] b. For each month, the prices of each stock can be displayed in matrices.

To find the total value of Mrs. Rivera’s portfolio for each month, multiply the the matrices.

matrix found in part a by each of

The total value of Mrs. Rivera’s portfolio for each month is $4379.64 in July, $4019.65 in August, and $3254.83 in September.

95. SAT/ACT A circle is inscribed in a square and intersects the square at points A, B, C, and D. If AC = 12, what is the total area of the shaded regions? eSolutions Manual - Powered by Cognero

Page 56

totalof value of Mrs. Rivera’s portfolio for each month is $4379.64 in July, $4019.65 in August, and $3254.83 in 2-4 The Zeros Polynomial Functions September.

95. SAT/ACT A circle is inscribed in a square and intersects the square at points A, B, C, and D. If AC = 12, what is the total area of the shaded regions?

A 18 B 36 C 18 D 24 E 72

SOLUTION:   The shaded regions located above AC can be reflected in AC to completely fill the region below AC.

2 Since AC = 12, the sides of the square are 12 units in length. Thus, the total area of the square is 12 or 144 units. Since half of the square is shaded, the area of the shaded region is

or 72.

The correct answer is E.

96. REVIEW f (x) = x2 – 4x + 3 has a relative minimum located at which of the following x-values? F –2 G2 H3 J4

SOLUTION:   Use a graphing calculator to graph the function.

97. Find all of the zeros of p (x) = x3 + 2x2 – 3x + 20.           A –4, 1 + 2i, 1 – 2i           B – 1, 1, 4 + i, 4 – i           C 1, 4 + i, 4 – i           D 4, 1 + i, 1 – i

SOLUTION:   eSolutions Manual - Powered by Cognero

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 20. Therefore, the possible rational zeros of g are ±1, ±2, ±4, ±5, ±10, and ±20. By using synthetic division, it can be determined that x = −4 is a rational zero.

Page 57

2-4 Zeros of Polynomial Functions 97. Find all of the zeros of p (x) = x3 + 2x2 – 3x + 20.           A –4, 1 + 2i, 1 – 2i           B – 1, 1, 4 + i, 4 – i           C 1, 4 + i, 4 – i           D 4, 1 + i, 1 – i

SOLUTION:   Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 20. Therefore, the possible rational zeros of g are ±1, ±2, ±4, ±5, ±10, and ±20. By using synthetic division, it can be determined that x = −4 is a rational zero.

2

The remaining quadratic factor (x − 2x + 5) yields no real zeros. Use the quadratic formula to find the zeros.

Thus, the zeros of p are −4, 1 + 2i, and 1 − 2i. The correct answer is A.

98. REVIEW Which expression is equivalent to (t2 + 3t – 9)(5 – t)–1? F G –t – 8 H –t – 8 + J

SOLUTION:   2

(t + 3t – 9)(5 – t)

–1

2

can be written as (t + 3t – 9) ·

The quotient is eSolutions Manual - Powered by Cognero

or

.

. The correct answer is H. Page 58

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