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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)