EXAMPLE 4

1. 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. SOLUTION f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8. The coefficients in f (x) have3 sign changes, so f has 3or 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

2. The possible numbers of zeros for f are summarized in the table below. EXAMPLE 4 Use Descartes’ rule of signs The coefficients in f (– x) have3sign changes, so f has 3 or 1 negative real zero(s) .

3. for Example 4 GUIDED PRACTICE Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function. 9. f (x) = x3 + 2x –11 SOLUTION f (x) = x3 + 2x – 11 The coefficients in f (x) have1 sign changes, so f has 1 positive real zero(s).

4. The possible numbers of zeros for f are summarized in the table below. for Example 4 GUIDED PRACTICE f (– x) = (– x)3 + 2(–x) – 11 = – x3 – 2x – 11 The coefficients in f (– x) haveno sign changes.

5. for Example 4 GUIDED PRACTICE 10. g(x) = 2x4 – 8x3 + 6x2 – 3x + 1 SOLUTION f (x) = 2x4 – 8x3 + 6x2 – 3x + 1 The coefficients in f (x) have4 sign changes, so f has 4 positive real zero(s). f (– x) = 2(– x)4 – 8(–x)3 + 6(– x)2 + 1 = 2x4 + 8x + 6x2 + 1 The coefficients in f (– x) haveno sign changes.

6. The possible numbers of zeros for f are summarized in the table below. for Example 4 GUIDED PRACTICE