Answer
$f(x)=x^3-5x^2+3x+9$
Work Step by Step
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 $-1$ (multiplicity 1) and $3$ (multiplicity 2), hence
$f(x)=a(x+1)(x-3)^2=a(x^3-5x^2+3x+9)$
If $a=1$, the function becomes
$f(x)=x^3-5x^2+3x+9$