Answer
(a) $\frac{p}{q}=\pm1,\pm\frac{1}{2},\pm3,\pm\frac{3}{2}$
(b) $P(x)=(x-3)(2x-1)(x+1)$
(c) $-1, \frac{1}{2}, 3$
(d) see plot.
Work Step by Step
We have $P(x)=2x^3-5x^2-4x+3$
(a) $p=\pm1,\pm3,q=\pm1,\pm2$,
possible rational zeros: $\frac{p}{q}=\pm1,\pm\frac{1}{2},\pm3,\pm\frac{3}{2}$
(b) Use synthetic division, find one zero of P and the quadratic form:
$P(x)=(x-3)(2x^2+x-1)=(x-3)(2x-1)(x+1)$
(c) The zeros of P are $-1, \frac{1}{2}, 3$
(d) The y-intercept is $P(0)=3$.
End behaviors: when $x\to -\infty, y\to -\infty$; when $x\to \infty, y\to \infty$
Test points $P(1)=-4,P(2)=-9$
Use the above information to plot the function.
