Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 284: 81

$-2,-1,\frac{1}{2},1$

The possible rational zeros are $\pm1,\pm1/2,\pm2$ Use synthetic division or the remainder theorem to test each value and we can confirm two zeros as $-1,+1$ $P(1)=2+3-4-3+2=0$ $P(-1)=2-3-4+3+2=0$ Again, use synthetic division to reduce P into a quadratic form as $P(x)=(x+1)(x-1)(2x^2+3x-2)=(x-1)(x+1)(2x-1)(x+2)$ so the zeros are $-2,-1,1/2,1$