2. And the zeros are x =  3, x = , and x = 2

3. Since the remainder is –64, we know that x + 3 is not a factor. Quotient Divisor Dividend

5.  

8. Use synthetic division to find the quotient and remainder. The quotient is – 4x 4 – 7x 3 – 8x 2 – 14x – 28 and the remainder is –6, also f (2) =  6. –6–28–14–8–7–4 –56–28–16–14–8 500261–42 Note: We must write a 0 for the missing term.

10. (a)List all possible rational zeros. (b)Find all rational zeros and factor P(x).

12. Example 6 Determine the possible number of positive real zeros and negative real zeros of P(x) = x 4 – 6x 3 + 8x 2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. P(x) = +x 4 – 6x 3 + 8x 2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(  x). P(  x) = (  x) 4 – 6(  x) 3 + 8(  x) 2 + 2(  x)  1 = x 4 + 6x 3 + 8x 2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root. 1 2 3

14. Example 7 Show that the real zeros of P(x) = 2x 4 – 5x 3 + 3x + 1 satisfy the following conditions. (a) No real zero is greater than 3. (b) No real zero is less than –1. Solution a) c > 0 b) c < 0 All are nonegative. The numbers alternate in sign. No real zero greater than 3. No zero less than  1.