Answer
$P(x)=-2x^4+4x^3+10x^2-12x$
Work Step by Step
RECALL:
If c is a zero of a polynomial, then $x-c$ is a factor of the polynomial.
Since -2, 0, 1, and 3 are zeros of the polynomial, then
(x+2),(x), (x-1), and (x−3) are factors of the polynomial.
Thus, the polynomial of degree 4 with the given zeros is:
$\\P(x)=a(x+2)(x)(x−1)(x−3)
\\P(x)=a(x^4-2x^3-5x^2+6x)$
where $a$ is a real number.
Since the coefficient of $x^3$ is 4, then $a$ must be:
$a(-2x^3)=4x^3
\\-2ax^3=4x^3
\\-2a=4
\\a=\frac{4}{-2}
\\a=-2$
Thus, the polynomial with the given zeros whose coefficient of $x^3$ is 4 is:
$P(x)=-2(x^4-2x^3-5x^2+6x)
\\P(x)=-2x^4+4x^3+10x^2-12x$
