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.
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.
By using synthetic division on the depressed polynomial, it can be determined that x = 6 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.
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 .
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By using synthetic division, it can be determined that
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is a rational zero.
2-4
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 −of1)(x
5). Factoring the quadratic expression yields f (x) = (x – 1)(x – 6)(x + 5)(x + 1). Thus,
− 6)(x +6x +Functions
Zeros
Polynomial
the rational zeros of g are 1, 6, −5, and −1.
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, it can be determined that
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
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.
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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 − 12) yields no rational zeros, the rational zeros of h 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 depressed polynomial, it can be determined that x = 4 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).
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.
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2
The depressed polynomial 2x + x + 9 has no real zeros. Thus, h = 5. The dimensions of the container are 5, 5 −4 or
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
Because the factor (x + 9) yields no real zeros, the rational zeros of h are 3 and 4
− 4) (x + 9).Functions
Zeros
Polynomial
(multiplicity: 2).
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.
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).
4
13. 3
2
x –Manual
3x –- 20x
+ 84x
– 80 = 0
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by Cognero
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SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is
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
2-4
2
(x − 1)(x + 4)(x + 9x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x + 4)(x + 3) . Thus, the solutions
Zeros
Polynomial
are
1, −4,of
and
2).
−3 (multiplicity:Functions
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.
By using synthetic division on the depressed polynomial, it can be determined that x = −5 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).
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
By using synthetic division, it can be determined that x =
.
is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x =
Because
and 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
,
is a rational zero.
, and −4 (multiplicity: 2).
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.Thus,
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17. SALES The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) =
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
+ 5)(x
the quadratic expression yields 0 = (x − 4)(x + 5)(x − 2)
− 4)(xof
− 4x + 4). Factoring
2-4 (x
Zeros
Polynomial
Functions
. Thus, the solutions
are 4, −5, and 2 (multiplicity: 2).
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
.
By using synthetic division, it can be determined that x =
is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x =
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).
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.
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. Factoring the quadratic expression yields
.Thus,
2-4 the
Zeros
of Polynomial
solutions
are , , and Functions
−4 (multiplicity: 2).
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.
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