## Precalculus (10th Edition)

$f(x)=x^3-12x-16$
If $c$ is a zero of a function with multiplicity $b$ then $(x-c)^b$ is a “factor” of the function. We are given that the degree is $3$, and the zeros are $-2$ (multiplicity 2) and $4$ (multiplicity 1), hence $f(x)=a(x+2)^2(x-4)\\ f(x)=a(x^3-12x-16)$ If $a=1$, the function becomes $f(x)=x^3-12x-16$