í )( - Montville Township Public Schools / Overvie all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 ± 6x3 ± 31 x2 + 216 x í 180

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

(x) = (x 1)(x 5)(x2 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:

The leading coefficient is 4 and the constant term is 6. The possible rational zeros are or

.

By using synthetic division, it can be determined that x = 6 is a rational zero.

Because (x 6) is a factor of f (x), we can write a factored form of f (x) as f (x) = (x 6)(4x2 1). To find zeros of

4x2 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 . By using synthetic division, it can be determined that x = 1 is a rational zero.



(x) = (x 1)(x 6)(x2 +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:


.




(x) = (x 2)(x 5)(4x2 +7x + 2). Factoring the quadratic expression yields f (x) = (x 2)(x 5)(x 2)(4x 1) or

(x 2)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

.

By using synthetic division, it can be determined that isarationalzero.

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

Because and arefactorsofh(x), we can use the final quotient to write a factored form of h(x) as

Factoringthequadraticexpressionyields

Becausethefactor(x2 12) yields no rational zeros, the rational zeros of h are

6.f (x) = 18x4 + 12x3 + 56x2 + 48x 64

SOLUTION:


or .



Because and arefactorsoff (x), we can use the final quotient to write a factored form of f (x) as


Becausethefactor(x2 + 4) yields no real zeros, the rational zeros of f are

7.h(x) = x5 11x4 + 49x3 147x2 + 360x 432

SOLUTION:


.



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

Because (x 3) and (x 4)2 are factors of h(x), we can use the final quotient to write a factored form of h(x) as h

(x) = (x 3)(x 4)2(x2 + 9). Because the factor (x2 + 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:


.

. By using synthetic division, it can be determined that isarationalzero.

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

Because and arefactorsofg(x), we can use the final quotient to write a factored form of f (x) as

Tofindzerosof8x3 72, we can set it equal to zero and solve for x.

Because the factor (8x3 72) yields no rational zeros, the rational zeros of g are

9.MANUFACTURINGThespecificationsforthedimensionsofanewcardboardcontainerareshown.Ifthevolume of the container is modeled by V(h) = 2h

3 9h2 + 4h and it will hold 45 cubic inches of merchandise, what

are the container's dimensions?

SOLUTION:

Substitute V(h) = 45 into V(h) = 2h3 9h

2 + 4h and apply the Rational Zeros Theorem to find possible rational zeros

of the function.


.

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

The depressed polynomial 2x2 + 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 =

(x + 1)(x 3)(x2 + 4x + 4). Factoring the quadratic expression yields 0 = (x + 1)(x 3)(x + 2)2. Thus, the solutions are 1, 3, and 2 (multiplicity: 2).

11.x4 + 9x3 + 23x2 + 3x 36 = 0


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.


Because (x 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x 1)(x + 4)(x2 + 9x + 9). Factoring the quadratic expression yields 0 = (x 1)(x + 4)(x + 3)2. 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.



(x + 1)(x 2)(x2 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





(x 4)(x + 5)(x2 4x + 4). Factoring the quadratic expression yields 0 = (x 4)(x + 5)(x 2)2. Thus, the solutions are 4, 5, and 2 (multiplicity: 2).

14.x4 + 34x = 6x3 + 21x2 48

SOLUTION:

The equation can be written as x4 6x3 21x2 + 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 .




(x + 1)(x 2)(x2 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:

The equation can be written as 6x4 + 41x

3 + 42x

2 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 = isarationalzero.


Because and arefactorsoftheequation,wecanusethefinalquotienttowriteafactoredformas

. Factoring the quadratic expression yields .Thus,

the solutions are , , and 4 (multiplicity: 2).

16.12x4 + 77x3 = 136x2 33x 18

SOLUTION:


3 136x2 + 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 .




. Factoring the quadratic expression yields .

Thus, the solutions are

17.SALESThesalesS(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 2x

3 2x2 + 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:

Substitute S(x) = 16 into S(x) = 2x3 2x

2 + 4x and apply the Rational Zeros Theorem to find possible rational zeros

of the function.

The equation can be written as 2(x3 x2 + 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.

The depressed polynomial x2 + 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 listto 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

(x) = (x + 2)(x 1)(x2 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




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

constant term is 24. The possible rational zeros are or1, 2, 3, 4, 6, 8, 12,

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 arereasonable.



Because (x + 4) and arefactorsoftheequation,wecanusethefinalquotienttowriteafactoredformas

. Factoring the quadratic expression yields

. 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].




constant term is 16. The possible rational zeros are or1, 2, 4, 8, 16, and . Because the

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

(x) = (x + 4)(x + 1)(2x2 6x + 4). Factoring the quadratic expression yields g(x) = 2(x + 4)(x + 1)(x 2)(x 1).


21.g(x) = 6x4 33x3 6x2 + 123x 90




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 theconstant 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.



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

(x) = (x + 2)(x 1)(6x2 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





constant term is 20. The possible rational zeros are or1, 2, 4, 5, 10, 20, , and

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

graph, it appears that arereasonable.



Because and(x 5) are factors of the equation, we can use the final quotient to write a factored form as


Thus, the solutions are .

23.f (x) = 2x4 13x3 + 21x2 + 9x 27





constant term is 27. The possible rational zeros are or1, 3, 9, 27, , , and .

Because the real zeros are in the interval [2, 7], narrow this list to just 1, 3, , , and . From the graph, it

appears that 1 and arereasonable.





Thus, the solutions are 1, , 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].




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

form as h(x) = (x + 2)(x 1)(x 3)(x2 + x 2). Factoring the quadratic expression yields f (x) = (x + 2)(x 1)(x

3)(x + 2)(x 1) or (x + 2)2(x 1)

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

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





constant term is 18. The possible rational zeros are or1, 3, 6, 9, 18, , , ,

, , 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, and3arereasonable.




Because arefactorsoftheequation,wecanusethefinalquotienttowriteafactored

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). f(x) = 2x3 3x2 + 4x + 7 f(x) has one variation in sign.

f(x) has 2 variations in sign. By DescartesRule 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). f(x) = 10x

4 3x3 + 8x2 4x 8

f(x) has three variations in sign.

f(x) has 1 variation in sign. By DescartesRule 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). f(x) = 3x4 5x3 + 4x2 + 2x 6 f(x) has two variations in sign.

f(x) has two variations in sign. By DescartesRule 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


4 + 6x

3 + 3x

2 2x + 12

f(x) has two variations in sign.


30.g(x) = 4x5 + 3x4 + 9x3 8x2 + 16x 24

SOLUTION:Examine the variations in sign for g(x) and for g(x). g(x) = 4x

5 + 3x

4 + 9x

3 8x2 + 16x 24

g(x) has three variations in sign.

g(x) has two variations in sign. By DescartesRule 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). h(x) = 4x5 + x4 8x3 24x2 + 64x 124 h(x) has four variations in sign.

h(x) has one variation in sign. By DescartesRule 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.

Therefore, a function of least degree that has 3, 4, 6, and 1 as zeros is f (x) = x4 4x

3 23x

2 + 54x + 72 or any

nonzero multiple of f (x).

33.2, 4, 3, 5

SOLUTION:


Therefore, a function of least degree that has 2, 4, 3, and 5 as zeros is f (x) = x4 + 4x3 19x2 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.

Therefore, a function of least degree that has 5, 3, 4 + i, and 4 i as zeros is f (x) = x4 6x

3 14x

2 + 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.

Therefore, a function of least degree that has 1, 8, 6 i, and 6 + i as zeros is f (x) = x4 19x3 + 113x2 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 , 2 , 3, and 7 as zeros is f (x) = x4 4x3 41x2 + 80x + 420or any nonzero multiple of f (x).

37.5, 2, 4 , 4 +

SOLUTION:

Using the Linear Factorization Theorem and the zeros 5, 2, 4 , and 4 + , write f (x) as follows.

f(x) = a[x (5)][x (2)][x (4 )][x (4 + )]


Therefore, a function of least degree that has 5, 2, 4 , and 4 + aszerosisf (x) = x4 5x3 21x2 + 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)]


Therefore, a function of least degree that has , , 4i, and 4i as zeros is f (x) = x4 + 9x2 112 or any nonzero multiple of f (x).

39. , , 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)]


Therefore, a function of least degree that has , , 3 4i, and 3 + 4i as zeros is f (x) = x4 6x3 + 19x2 + 36x 150 or any nonzero multiple of f (x).

40.2 + , 2 , 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)]


Therefore, a function of least degree that has 2 + , 2 , 4 + 5i, and 4 5i as zeros is f (x) = x4 12x3 + 74x2

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)]


Therefore, a function of least degree that has 6 , 6 + , 8 3i, and 8 + 3i as zeros is f (x) = x4 28x3 +

296x2 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 canbe determined that x = 4 is a rational zero.


The remaining quadratic factor (x2 + 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)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.


The remaining quadratic factor (x2 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 + )(x 3 ).

c. The zeros are 2, 1, 3 , 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.


The remaining quadratic factor (x2 + 9) yields no real zeros and is therefore, irreducible over the reals. So, h(x)

written as a product of linear and irreducible quadratic factors is h(x) = (x2 + 9)(x 4)(x + 6).

b. x2 + 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

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.


The remaining quadratic factor (4x2 16x + 52) can be written 4(x

2 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

or (x2 4x + 13)(4x 3)(x 4)

b. Use the quadratic formula to find the zeros of x2 4x + 13.

x2 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.




or (4x + 5)(x + 3)(x2 8x + 41)



(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.


The remaining quadratic factor (x2 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 + )(x ).

c. The zeros are 3, 5, , and .

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

180. By using synthetic division, it can be determined that g(x) has no rational zeros. Substitute u = x2 to factor g(x).

x2 5 can be written as (x + )(x ). The remaining quadratic factor (x2 + 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 )(x2 + 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 )(x + 6i)(x 6i).

c. The zeros are , , 6i, and 6i.

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 verify 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.

Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 3i)(x 3i)(2x3 + x2

25x + 12). 2x3 + x

2 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.

The remaining depressed polynomial 2x2 +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 verify that 5i is a zero of h(x).


Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 5i)(x 5i)(3x3 5x

2

88x + 60). 3x3 5x2 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.

The remaining depressed polynomial 3x2 20x + 12 can be written as (x 6)(3x 2). The zeros of the depressed


(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 verify 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.

Using these two zeros and the depressed polynomial from the last division, write h(x) = [x (3 i)][x (3 + i)](x3 +

4x2 + x 6). x3 + 4x2 + x 6 has possible rational zeros of 1, 2, 3, and 6. By using synthetic division, it can be


The remaining depressed polynomial x2 + 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 verify 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)](4x3

25x2 + 19x + 30). 4x3 25x2 + 19x + 30 has possible rational zeros of 1, 2, 3, 5, 6, 10, 15, 30,

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

zero.

The remaining depressed polynomial 4x2 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 . 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 verify 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.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x 2i)(x3 3x

2 8x

+ 24). x3 3x2 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.

The remaining depressed polynomial x2 8 can be written as (x )(x + ) or (x 2 )(x + 2 ). The

zeros of the depressed polynomial are 2 and 2 . Therefore, the zeros of f are 3, 2 , 2 , 2i, and 2i.

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

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



Using these two zeros and the depressed polynomial from the last division, write g(x) = [x (3 + i)][x (3 i)](x2

4x + 1). x2 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.ARCHITECTUREAnarchitectisconstructingascalemodelofabuildingthatisintheshapeofapyramid. 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 terms 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

sides of the square base. The area of the base B is B = l2. The height h is h = 9. Substitute these values for B and h into the formula for the volume of a pyramid.

A polynomial function that describes the volume of the model in terms of its length is V(l) = l3 3 2.

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

6300 = l3 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 =30isarealzero.

Use synthetic division to test this possibility.

mustbeapositivevalue.BecauseV( ) only has 1 positive real zero, no other tests are necessary. Thus, the baseis 30 inches by 30 inches and the height of the model is 21 inches.

56.CONSTRUCTIONTheheightofatunnelthatisunderconstructionis1footmorethanhalfitswidthanditslength 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 + andthewidthisw = 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.

The polynomial 81w3 + 170w

2 + 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 + or37feet,andthelengthis32+324(72)or23,360feet.

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 , write f (x) as follows.


Since is a factor, when .

A polynomial function of least degree with integer coefficients that has asazeroisf (x) = x3 6.

58.

SOLUTION:



. Since isafactor, when .

A polynomial function of least degree with integer coefficients that has asazeroisf (x) = x3 5.

59.

SOLUTION:




A polynomial function of least degree with integer coefficients that has asazeroisf (x) = x3 + 2.

60.

SOLUTION:




A polynomial function of least degree with integer coefficients that has asazeroisf (x) = x3 + 7.

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.

The remaining quadratic factor (3x2 + 75) can be written as 3(x

2 + 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.

The remaining factor (2x3 + 32x) can be written as 2x(x

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.

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 or1,

2, 3, 4, 6, 12, , , , , , and .


The remaining quadratic factor (6x2 12x + 12) can be written as 6(x

2 2x + 2). x2 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.


The remaining quadratic factor (x2 + 4) yields no real zeros. Thus, the rational zeros of f are 4 and 2.

65.w(z) = z4 10z3 + 30z2 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 verify 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

and the constant term is 1. The possible rational zeros are or1, , , and .



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

The remaining quadratic factor (6x2 + 6) can be written as 6(x

2 + 1) and yields no real zeros. Thus, the rational

zeros of b are

67.ENGINEERINGAsteelbeamissupportedbytwopilings200feetapart.Ifaweightisplacedx feet from the piling on the left, a vertical deflection represented by d = 0.0000008x

2(200 x) occurs. How far is the weight from

the piling if the vertical deflection is 0.8 feet?

SOLUTION:

Solve d = 0.0000008x2(200 x) for d = 0.8.

Use a graphing calculator to graph y = 0.0000008x2(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:

Use the difference of two cubes, a3 b

3 = (a b)(a

2 + ab + b

2), to factor x

3 3.

69.x3 + 16

SOLUTION:

Use the sum of two cubes, a3 + b

3 = (a + b)(a

2 ab + b2), to factor x

3 + 16.

70.8x3 + 9

SOLUTION:


3 = (a + b)(a

2 ab + b2), to factor 8x

3 + 9.

71.27x6 + 4

SOLUTION:


3 = (a + b)(a

2 ab + b2), to factor 27x

6 + 4.

72.MULTIPLEREPRESENTATIONS In this problem, you will explore even- and odd-degree polynomial functions. a. ANALYTICALIdentifythedegreeandnumberofzerosofeachpolynomialfunction.

i. f (x) = x3 x2 + 9x 9

ii. g(x) = 2x5 + x

4 32x 16

iii. h(x) = 5x3 + 2x2 13x + 6

iv. f (x) = x4 + 25x

2 + 144

v. h(x) = 3x6 + 5x5 + 46x4 +80x3 32x2

vi. g(x) = 4x4 11x

3 + 10x

2 11x + 6

b. NUMERICALFindthezerosofeachfunction. c. VERBALDoesanodd-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.

The remaining quadratic factor (x2 + 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 or1,

2, 6, 8, 16, and .



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

The remaining quadratic factor (2x2 + 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 or1, 2,

3, 6, .


The remaining quadratic factor (5x2 + 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).

eSolutions Manual - Powered by Cognero Page 1

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:


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



(x) = (x 1)(x 5)(x2 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:


.


Because (x 6) is a factor of f (x), we can write a factored form of f (x) as f (x) = (x 6)(4x2 1). To find zeros of

4x2 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 . By using synthetic division, it can be determined that x = 1 is a rational zero.



(x) = (x 1)(x 6)(x2 +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:


.




(x) = (x 2)(x 5)(4x2 +7x + 2). Factoring the quadratic expression yields f (x) = (x 2)(x 5)(x 2)(4x 1) or

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

5.h(x) = 6x4 + 13x3 67x2 156x 60

SOLUTION:


or

.



Because and arefactorsofh(x), we can use the final quotient to write a factored form of h(x) as


Becausethefactor(x2 12) yields no rational zeros, the rational zeros of h are

6.f (x) = 18x4 + 12x3 + 56x2 + 48x 64

SOLUTION:


or .



Because and arefactorsoff (x), we can use the final quotient to write a factored form of f (x) as


Becausethefactor(x2 + 4) yields no real zeros, the rational zeros of f are

7.h(x) = x5 11x4 + 49x3 147x2 + 360x 432

SOLUTION:


.



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

Because (x 3) and (x 4)2 are factors of h(x), we can use the final quotient to write a factored form of h(x) as h

(x) = (x 3)(x 4)2(x2 + 9). Because the factor (x2 + 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:


.

. By using synthetic division, it can be determined that isarationalzero.


Because and arefactorsofg(x), we can use the final quotient to write a factored form of f (x) as

Tofindzerosof8x3 72, we can set it equal to zero and solve for x.

Because the factor (8x3 72) yields no rational zeros, the rational zeros of g are

9.MANUFACTURINGThespecificationsforthedimensionsofanewcardboardcontainerareshown.Ifthevolume of the container is modeled by V(h) = 2h

3 9h2 + 4h and it will hold 45 cubic inches of merchandise, what

are the container's dimensions?

SOLUTION:

Substitute V(h) = 45 into V(h) = 2h3 9h

2 + 4h and apply the Rational Zeros Theorem to find possible rational zeros

of the function.


.

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

The depressed polynomial 2x2 + 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


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.



(x + 1)(x 3)(x2 + 4x + 4). Factoring the quadratic expression yields 0 = (x + 1)(x 3)(x + 2)2. Thus, the solutions are 1, 3, and 2 (multiplicity: 2).

11.x4 + 9x3 + 23x2 + 3x 36 = 0





(x 1)(x + 4)(x2 + 9x + 9). Factoring the quadratic expression yields 0 = (x 1)(x + 4)(x + 3)2. 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.



(x + 1)(x 2)(x2 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





(x 4)(x + 5)(x2 4x + 4). Factoring the quadratic expression yields 0 = (x 4)(x + 5)(x 2)2. Thus, the solutions are 4, 5, and 2 (multiplicity: 2).

14.x4 + 34x = 6x3 + 21x2 48

SOLUTION:

The equation can be written as x4 6x3 21x2 + 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 .




(x + 1)(x 2)(x2 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:


3 + 42x

2 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 .




. Factoring the quadratic expression yields .Thus,

the solutions are , , and 4 (multiplicity: 2).

16.12x4 + 77x3 = 136x2 33x 18

SOLUTION:


3 136x2 + 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 .





Thus, the solutions are

17.SALESThesalesS(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 2x

3 2x2 + 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:

Substitute S(x) = 16 into S(x) = 2x3 2x

2 + 4x and apply the Rational Zeros Theorem to find possible rational zeros

of the function.

The equation can be written as 2(x3 x2 + 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.

The depressed polynomial x2 + 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





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 listto 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

(x) = (x + 2)(x 1)(x2 10x + 24). Factoring the quadratic expression yields f (x) = (x + 2)(x 1)(x 4)(x 6).


19.f (x) = 2x4 x3 29x2 + 34x + 24





constant term is 24. The possible rational zeros are or1, 2, 3, 4, 6, 8, 12,

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 arereasonable.




. Factoring the quadratic expression yields

. Thus, the solutions are

20.g(x) = 2x4 + 4x3 18x2 4x + 16





constant term is 16. The possible rational zeros are or1, 2, 4, 8, 16, and . Because the

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

(x) = (x + 4)(x + 1)(2x2 6x + 4). Factoring the quadratic expression yields g(x) = 2(x + 4)(x + 1)(x 2)(x 1).


21.g(x) = 6x4 33x3 6x2 + 123x 90




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 theconstant 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.



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

(x) = (x + 2)(x 1)(6x2 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





constant term is 20. The possible rational zeros are or1, 2, 4, 5, 10, 20, , and

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

graph, it appears that arereasonable.



Because and(x 5) are factors of the equation, we can use the final quotient to write a factored form as


Thus, the solutions are .

23.f (x) = 2x4 13x3 + 21x2 + 9x 27





constant term is 27. The possible rational zeros are or1, 3, 9, 27, , , and .

Because the real zeros are in the interval [2, 7], narrow this list to just 1, 3, , , and . From the graph, it

appears that 1 and arereasonable.





Thus, the solutions are 1, , and 3 (multiplicity: 2).

24.h(x) = x5 x4 9x3 + 5x2 + 16x 12





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.



Because (x + 2), (x 1), and (x 3) are factors of the equation, we can use the final quotient to write a factored

form as h(x) = (x + 2)(x 1)(x 3)(x2 + x 2). Factoring the quadratic expression yields f (x) = (x + 2)(x 1)(x

3)(x + 2)(x 1) or (x + 2)2(x 1)

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

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





constant term is 18. The possible rational zeros are or1, 3, 6, 9, 18, , , ,

, , 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, and3arereasonable.




Because arefactorsoftheequation,wecanusethefinalquotienttowriteafactored

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). f(x) = 2x3 3x2 + 4x + 7 f(x) has one variation in sign.

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

27.f (x) = 10x4 3x3 + 8x2 4x 8


4 3x3 + 8x2 4x 8

f(x) has three variations in sign.

f(x) has 1 variation in sign. By DescartesRule 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). f(x) = 3x4 5x3 + 4x2 + 2x 6 f(x) has two variations in sign.


29.f (x) = 12x4 + 6x3 + 3x2 2x + 12


4 + 6x

3 + 3x

2 2x + 12

f(x) has two variations in sign.


30.g(x) = 4x5 + 3x4 + 9x3 8x2 + 16x 24

SOLUTION:Examine the variations in sign for g(x) and for g(x). g(x) = 4x

5 + 3x

4 + 9x

3 8x2 + 16x 24

g(x) has three variations in sign.

g(x) has two variations in sign. By DescartesRule 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). h(x) = 4x5 + x4 8x3 24x2 + 64x 124 h(x) has four variations in sign.

h(x) has one variation in sign. By DescartesRule 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:


Therefore, a function of least degree that has 3, 4, 6, and 1 as zeros is f (x) = x4 4x

3 23x

2 + 54x + 72 or any

nonzero multiple of f (x).

33.2, 4, 3, 5

SOLUTION:


Therefore, a function of least degree that has 2, 4, 3, and 5 as zeros is f (x) = x4 + 4x3 19x2 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.

Therefore, a function of least degree that has 5, 3, 4 + i, and 4 i as zeros is f (x) = x4 6x

3 14x

2 + 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.

Therefore, a function of least degree that has 1, 8, 6 i, and 6 + i as zeros is f (x) = x4 19x3 + 113x2 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)]


Therefore, a function of least degree that has 2 , 2 , 3, and 7 as zeros is f (x) = x4 4x3 41x2 + 80x + 420or any nonzero multiple of f (x).

37.5, 2, 4 , 4 +

SOLUTION:

Using the Linear Factorization Theorem and the zeros 5, 2, 4 , and 4 + , write f (x) as follows.

f(x) = a[x (5)][x (2)][x (4 )][x (4 + )]


Therefore, a function of least degree that has 5, 2, 4 , and 4 + aszerosisf (x) = x4 5x3 21x2 + 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)]


Therefore, a function of least degree that has , , 4i, and 4i as zeros is f (x) = x4 + 9x2 112 or any nonzero multiple of f (x).

39. , , 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)]


Therefore, a function of least degree that has , , 3 4i, and 3 + 4i as zeros is f (x) = x4 6x3 + 19x2 + 36x 150 or any nonzero multiple of f (x).

40.2 + , 2 , 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)]


Therefore, a function of least degree that has 2 + , 2 , 4 + 5i, and 4 5i as zeros is f (x) = x4 12x3 + 74x2

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)]


Therefore, a function of least degree that has 6 , 6 + , 8 3i, and 8 + 3i as zeros is f (x) = x4 28x3 +

296x2 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 canbe determined that x = 4 is a rational zero.


The remaining quadratic factor (x2 + 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)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.


The remaining quadratic factor (x2 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 + )(x 3 ).

c. The zeros are 2, 1, 3 , 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.


The remaining quadratic factor (x2 + 9) yields no real zeros and is therefore, irreducible over the reals. So, h(x)

written as a product of linear and irreducible quadratic factors is h(x) = (x2 + 9)(x 4)(x + 6).

b. x2 + 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

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.




or (x2 4x + 13)(4x 3)(x 4)



(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





or (4x + 5)(x + 3)(x2 8x + 41)



(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.


The remaining quadratic factor (x2 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 + )(x ).

c. The zeros are 3, 5, , and .

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

180. By using synthetic division, it can be determined that g(x) has no rational zeros. Substitute u = x2 to factor g(x).

x2 5 can be written as (x + )(x ). The remaining quadratic factor (x2 + 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 )(x2 + 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 )(x + 6i)(x 6i).

c. The zeros are , , 6i, and 6i.

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 verify that 3i is a zero of h(x).


Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 3i)(x 3i)(2x3 + x2

25x + 12). 2x3 + x

2 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.

The remaining depressed polynomial 2x2 +7x 4 can be written as (x + 4)(2x 1). The zeros of the depressed


(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 verify that 5i is a zero of h(x).


Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 5i)(x 5i)(3x3 5x

2

88x + 60). 3x3 5x2 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.

The remaining depressed polynomial 3x2 20x + 12 can be written as (x 6)(3x 2). The zeros of the depressed


(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 verify 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.

Using these two zeros and the depressed polynomial from the last division, write h(x) = [x (3 i)][x (3 + i)](x3 +

4x2 + x 6). x3 + 4x2 + x 6 has possible rational zeros of 1, 2, 3, and 6. By using synthetic division, it can be


The remaining depressed polynomial x2 + 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



Using

Documents

í )( - Montville Township Public Schools / Overvie all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 ± 6x3 ± 31 x2 + 216 x í 180