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

GUIDED PRACTICE for Example 2

Find all real zeros of the function.

3. f (x) = x3 – 4x2 – 15x + 18

Factors of the constant term: + 1, + 2, + 3, + 6, + 9

Factors of the leading coefficient: + 1

Possible rational zeros: + , + + 11

31

61

Simplified list of possible zeros: –3, 1 , 6

SOLUTION

GUIDED PRACTICE for Example 2

4. f (x) + x3 – 8x2 + 5x+ 14

SOLUTION

List the possible rational zeros. The leading coefficient is 1 and the constant term is 14. So, the possible rational zeros are:

x = + , + , + 71

21

11

STEP 1

GUIDED PRACTICE for Example 2

STEP 2

1 1 –8 5 14Test x =1:

1 –7 –21 –7 –2 12

Test x = –1:

–1 1 –8 5 14

1 –9 14 0 –1 9 14

1 is not a zero.↑

–1 is a zero.↑

Test these zeros using synthetic division.

GUIDED PRACTICE for Example 2

Because –1 is a zero of f, you can write f (x) = (x + 1)(x2 – 9x + 14).

STEP 3

f (x) = (x + 1) (x2 – 9x + 14)

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

The zeros of f are –1, 2, and 7.

ANSWER

= (x + 1)(x + 4)(x – 7)