1. EXAMPLE 2 Find all real zeros of f (x) = x 3 – 8x 2 +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 = +, +, +, +, +, + 5 1 4 1 2 1 1 1 10 1 20 1 STEP 1 Find zeros when the leading coefficient is 1

2. EXAMPLE 2 STEP 2 Find zeros when the leading coefficient is 1 1 1 – 8 11 20 Test x =1 : 1 – 7 4 1 – 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.

3. EXAMPLE 2 Because –1 is a zero of f, you can write f (x) = (x + 1)(x 2 – 9x + 20). STEP 3 f (x) = (x + 1) (x 2 – 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

4. GUIDED PRACTICE for Example 2 Find all real zeros of the function. 3. f (x) = x 3 – 4x 2 – 15x + 18 Factors of the constant term: + 1, + 2, + 3, + 6, + 9 Factors of the leading coefficient: + 1 Possible rational zeros: +, + + 1 1 3 1 6 1 Simplified list of possible zeros: –3, 1, 6 SOLUTION

5. GUIDED PRACTICE for Example 2 4. f (x) + x 3 – 8x 2 + 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 = +, +, + 7 1 2 1 1 1 STEP 1

6. GUIDED PRACTICE for Example 2 STEP 2 1 1 –8 5 14 Test x =1 : 1 –7 –2 1 –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.

7. GUIDED PRACTICE for Example 2 Because –1 is a zero of f, you can write f (x) = (x + 1)(x 2 – 9x + 14). STEP 3 f (x) = (x + 1) (x 2 – 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)