Answer
$P(x)=-2x^4+3x^3+3x^2-2x$
Work Step by Step
RECALL:
If c is a zero of a polynomial, then $x-c$ is a factor of the polynomial.
Since -1, 0, 2 and $\frac{1}{2}$ are zeros of the polynomial, then
(x+1),(x), (x-2), and $(x−\frac{1}{2})$ are factors of the polynomial.
Thus, the polynomial of degree 4 with the given zeros is:
$\\P(x)=a(x+1)(x)(x−2)(x−\frac{1}{2})
\\P(x)=a(x^4-\frac{3}{2}x^3-\frac{3}{2}x^2+x)$
where $a$ is a real number.
Since the coefficient of $x^3$ is 3, then $a$ must be:
$a(-\frac{3}{2}x^3)=3x^3
\\-\frac{3a}{2}x^3=3x^3
\\-\frac{3a}{2}=3
\\a=3 \cdot \frac{-2}{3}
\\a=-2$
Thus, the polynomial with the given zeros whose coefficient of $x^3$ is 4 is:
$\\P(x)=-2(x^4-\frac{3}{2}x^3-\frac{3}{2}x^2+x)
\\P(x)=-2x^4+3x^3+3x^2-2x$
