Answer
$p(x)=-2x^4+3x^3+3x^2-2x$
Work Step by Step
Since the zeros are $-1,0,2,\frac{1}{2}$, we have $p(x)=A(x+1)x(x-2)(x-\frac{1}{2})$.
Write in general form:
$p(x)=\frac{A}{2}(x^2+x)(x-2)(2x-1)$
$p(x)=\frac{A}{2}(x^2+x)(2x^2-5x+2)$
$p(x)=\frac{A}{2}(2x^4-3x^3-3x^2+2x)$
$p(x)=Ax^4-\frac{3A}{2}x^3-\frac{3A}{2}x^2+Ax$
In order that the coefficient of $x^3$ is 3, it must be $-\frac{3A}{2}=3\to A=-2$.
Now, we have $p(x)=-2x^4+3x^3+3x^2-2x$.
