Download 2-4 Zeros of Polynomial Functions

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 ±1, ±2, ±3, ±5, ±6, ±10, ±15, and ±30. By using synthetic division, it can be determined that 1 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that 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
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=
±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± , ± , ± , ±
,± ,± ,± ,± ,±
,±
, ± , and ± .
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
=
±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± , ± , ± , ±
By using synthetic division, it can be determined that
,± ,± ,± ,± ,±
and , ± , and ± .
is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
Because
,±
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
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 1 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that −4 is a rational zero.
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Because (x − 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as (x
2
2
− 1)(x + 4)(x + 6x + 9) = 0. Factoring the quadratic expression yields (x − 1)(x + 4)(x + 3) = 0. Thus, the solutions
Factoring the quadratic expression yields
2
2-4 Because
Zeros of
thePolynomial
factor (x − 12)Functions
yields no rational zeros, the rational zeros of h are
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 1 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that −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 (x
2
2
− 1)(x + 4)(x + 6x + 9) = 0. Factoring the quadratic expression yields (x − 1)(x + 4)(x + 3) = 0. Thus, the solutions
are 1, −4, and −3 (multiplicity: 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 ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±40, and ±80. By using synthetic division, it can be determined that 4 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that −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 (x
2
2
− 4)(x + 5)(x − 4x + 4) = 0. Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2) = 0. 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
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zeros are
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= ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ±
,±
, and ± .
Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as (x
2
2
4)(x + 5)(x
Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2)
− 4x + 4) = 0. Functions
2-4 −Zeros
of Polynomial
= 0. 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
= ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ±
zeros are
,±
, and ± .
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 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
.Thus, the solutions are
,
, 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.
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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
2-4 Zeros of Polynomial Functions
.Thus, the solutions are
,
, 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.
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.
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].
Test a lower bound of c = −6 and an upper bound of c = 5.
Every number in the last line is alternately nonnegative and nonpositive, so −6 is a lower bound.
Every number in the last line is nonnegative, so 5 is an upper bound.
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Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the
constant term is 24. 2-4 Every
Zerosnumber
of Polynomial
in the last line Functions
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. or ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24,
The possible rational zeros are
, and
. Because the real zeros are in the interval [−6, 5], narrow this list to just ±1, ±2, ±3, ±4, −6,
, and
. From the
are reasonable.
graph, it appears that
By using synthetic division, it can be determined that −4 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
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
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].
Test a lower bound of c = −4 and an upper bound of c = 7.
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2-4 Zeros of Polynomial Functions
Test a lower bound of c = −4 and an upper bound of c = 7.
The numbers in the last line alternate signs, , so −4 is a lower bound.
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. or ±1,±2, ±3, ±5, ±6, ±9, ±10, ±15,
The possible rational zeros are
±18, ±30, ±45, ±90, ± , ± , ± , ± , ±
,±
,± ,± ,± ,±
, ± , and ± . Because the real zeros are in the interval [−4, 7], narrow this list to just ±1,±2, ±3, 5, 6, ± , ± , ± , ± , ± , ± , ± ,
±
, ± , and ± . From the graph, it appears that −2, 1,
, and 5 are reasonable.
By using synthetic division, it can be determined that −2 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that 1 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 2
g(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.
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
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5.
4
3
2
23. fZeros
(x) = 2xof– Polynomial
13x + 21x + 9xFunctions
– 27
2-4
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].
Test a lower bound of c = −2 and an upper bound of c = 7.
The numbers in the last line alternate signs, so −2 is a lower bound.
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
= ±1, ±3, ±9, ±27, ±
,±
,±
and ±
. Because the real zeros are in the interval [−2, 7], narrow this list to just ±1, 3, ±
From the graph, it appears that −1 and
,±
, and
. are reasonable.
By using synthetic division, it can be determined that −1 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
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Because (x + 1) and
is a rational zero.
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are factors of the equation, we can use the final quotient to write a factored form as 2-4 Zeros of Polynomial Functions
Because (x + 1) and
are factors of the equation, we can use the final quotient to write a factored form as . . Thus, the solutions are −1,
Factoring the quadratic expression yields
, and 3
(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].
Test a lower bound of c = −3 and an upper bound of c = 5.
The numbers in the last line alternate signs, so −3 is a lower bound.
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
constant term is 18. The possible rational zeros are
or ±1, ±3, ±6, ±9, ±18, ±
,±
,±
,±
,±
, and ±
. Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±3, ±
From the graph, it appears that −2,
,±
,
,±
,±
, and ±
.
and 3 are reasonable.
By using synthetic division, it can be determined that −2 is a rational zero.
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Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±3, ±
,±
,
,±
,±
, and ±
.
theof
graph,
it appears that
−2, and 3 are reasonable.
2-4 From
Zeros
Polynomial
Functions
By using synthetic division, it can be determined that −2 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
is a rational zero.
By using synthetic division on the new depressed polynomial, it can be determined that 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
, which simplifies to
Thus, the solutions are −2,
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.
(multiplicity: 2), and 3 (multiplicity: 2).
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