EXAMPLE 5

Use the result to write f (x) as a product of two factors. Then factor completely.

f (x) = x3 – 2x2 – 23x + 60

The zeros are 3, – 5, and 4.

Standardized Test Practice

The correct answer is A. ANSWER

= (x – 3)(x + 5)(x – 4)

= (x – 3)(x2 + x – 20)

EXAMPLE 1 List possible rational zeros

List the possible rational zeros of f using the rational zero theorem.

a. f (x) = x3 + 2x2 – 11x + 12

Factors of the constant term: + 1, + 2, + 3, + 4, + 6, + 12

Factors of the leading coefficient: + 1

Possible rational zeros: + , + , + , + , + , +

11

21

31

41

61

121

Simplified list of possible zeros: + 1, + 2, + 3, + 4, + 6, + 12

EXAMPLE 1 List possible rational zeros

b. f (x) = 4x4 – x3 – 3x2 + 9x – 10

Factors of the constant term: + 1, + 2, + 5, + 10

Factors of the leading coefficient: + 1, + 2, + 4

+ , + , + , + , + , + , + , + , + , + , +

Possible rational zeros:11

21

51

101

12

22

52

102

14

24

54

+ 104

Simplified list of possible zeros: + 1, + 2, + 5, + 10, + , + , +

52

14

125

4+

EXAMPLE 2

Find all real zeros of f (x) = x3 – 8x2 +11x + 20.

SOLUTION

List the possible rational zeros. The leading coefficient is 1 and the constant term is 20. So, the possible rational zeros are:

x = + , + , + , + , + , +51

41

21

11

101

201

STEP 1

Find zeros when the leading coefficient is 1

EXAMPLE 2

STEP 2

Find zeros when the leading coefficient is 1

1 1 – 8 11 20Test x =1:

1 – 7 41 – 7 4 24

Test x = –1:

–1 1 –8 11 20

1 – 9 20 0 –1 9 20

1 is not a zero.↑

–1 is a zero↑

Test these zeros using synthetic division.

EXAMPLE 2

Because –1 is a zero of f, you can write f (x) = (x + 1)(x2 – 9x + 20).

STEP 3

f (x) = (x + 1) (x2 – 9x + 20)

Factor the trinomial in f (x) and use the factor theorem.

The zeros of f are –1, 4, and 5.

ANSWER

= (x + 1)(x – 4)(x – 5)

Find zeros when the leading coefficient is 1

EXAMPLE 1 Find the number of solutions or zeros

a. How many solutions does the equation x3 + 5x2 + 4x + 20 = 0 have?

SOLUTION

Because x3 + 5x2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)

EXAMPLE 1 Find the number of solutions or zeros

b. How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27 have?

SOLUTION

Because f (x) = x4 – 8x3 + 18x2 – 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)

EXAMPLE 2

Find all zeros of f (x) = x5 – 4x4 + 4x3 + 10x2 – 13x – 14.

SOLUTION

STEP 1 Find the rational zeros of f. Because f is a polynomial function of degree 5, it has 5 zeros. The possible rational zeros are + 1, + 2, + 7, and + 14. Using synthetic division, you can determine that – 1 is a zero repeated twice and 2 is also a zero.

STEP 2 Write f (x) in factored form. Dividing f (x) by its known factors x + 1, x + 1, and x – 2 gives a quotient of x2 – 4x + 7. Therefore:

f (x) = (x + 1)2(x – 2)(x2 – 4x + 7)

Find the zeros of a polynomial function

EXAMPLE 2

STEP 3 Find the complex zeros of f . Use the quadratic formula to factor the trinomial into linear factors.

f(x) = (x + 1)2(x – 2) x – (2 + i 3 ) x – (2 – i 3 )

The zeros of f are – 1, – 1, 2, 2 + i 3 , and 2 – i 3.

ANSWER

Find the zeros of a polynomial function

EXAMPLE 4 Use Descartes’ rule of signs

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.

f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.

The coefficients in f (x) have 3 sign changes, so f has 3 or 1 positive real zero(s).

f (– x) = (– x)6 – 2(– x)5 + 3(– x)4 – 10(– x)3 – 6(– x)2 – 8(– x) – 8

= x6 + 2x5 + 3x4 + 10x3 – 6x2 + 8x – 8

SOLUTION

EXAMPLE 4 Use Descartes’ rule of signs

The coefficients in f (– x) have 3 sign changes, so f has 3 or 1 negative real zero(s) .

The possible numbers of zeros for f are summarized in the table below.

EXAMPLE 5 Approximate real zeros

Approximate the real zeros of f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.

SOLUTION

Use the zero (or root) feature of a graphing calculator, as shown below.

From these screens, you can see that the zeros are x ≈ – 0.73 and x ≈ 2.73.

ANSWER

EXAMPLE 6 Approximate real zeros of a polynomial model

s (x) = 0.00547x3 – 0.225x2 + 3.62x – 11.0

What is the tachometer reading when the boat travels 15 miles per hour?

A tachometer measures the speed (in revolutions per minute, or RPMs) at which an engine shaft rotates. For a certain boat, the speed x of the engine shaft (in 100s of RPMs) and the speed s of the boat (in miles per hour) are modeled by

TACHOMETER

EXAMPLE 6 Approximate real zeros of a polynomial model

Substitute 15 for s(x) in the given function. You can rewrite the resulting equation as:

0 = 0.00547x3 – 0.225x2 + 3.62x – 26.0

Then, use a graphing calculator to approximate the real zeros of f (x) = 0.00547x3 – 0.225x2 + 3.62x – 26.0.

SOLUTION

From the graph, there is one real zero: x ≈ 19.9.

The tachometer reading is about 1990 RPMs.ANSWER