Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 284: 59

(a) $-1$ and $2$ (b) See graph.

(a) Possible rational zeros are $\pm1,\pm2,\pm4,\pm8$. Use synthetic division as shown in the figure, we can find two zeros at $x=-1,2$ and a quadratic so that $P(x)=(x+1)(x-2)(x^2-4x+4)=(x+1)(x-2)^3$, thus the zeros are $x=-1$ and $x=2$ with $x=2$ having a multiplicity of 3. (b) See graph of the equation and the zero in the figure.