## Precalculus: Mathematics for Calculus, 7th Edition

$P(x)=-2x^4+4x^3+10x^2-12x$
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: $P(x)=-2(x^4-2x^3-5x^2+6x) \\P(x)=-2x^4+4x^3+10x^2-12x$