Answer
$\color{blue}{\bf{f(x)=-3x^3+6x^2+33x-36}}$
Work Step by Step
We are given three zeros of a polynomial and the value of the function for a given ${x}$:
Our zeros are $\bf-3$, $\bf{1}$, and $\bf{4}$, so let's make them into factors of our polynomial.
If $\bf-3$ is a zero then:
$(-3+n)=0$
$n=3$
so our factor is $\bf{(x+3)}$ because when $x=\bf-3$, $(x+3)=0$
If $\bf1$ is a zero then
$(1+n)=0$
$n=-1$
so our factor is $\bf{(x-1)}$ because when $x=\bf1$, $(x-1)=0$
If $\bf4$ is a zero then
$(4+n)=0$
$n=-4$
so our factor is $\bf{(x-4)}$ because when $x=\bf4$, $(x-4)=0$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x+3)(x-1)(x-4)}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x+3)(x-1)(x-4)$
If $f(2)=30$ then
${(2+3)(2-1)(2-4)}$ times some number $a$, equals $30$ or:
$f(2)=30=a(2+3)(2-1)(2-4)$
$30=a(5)(1)(-2)$
$30=a(-10)$
$\dfrac{30}{-10}=a\dfrac{-10}{-10}$
$-3=a$
Now that we have the value of $a$, we can find $f(x)$
$f(x)=-3(x+3)(x-1)(x-4)$
$f(x)=-3(x+3)(x^2-5x+4)$
$f(x)=-3(x^3-2x^2-11x+12)$
$$\color{blue}{\bf{f(x)=-3x^3+6x^2+33x-36}}$$
