Answer
$$\color{blue}{\bf{f(x)= \dfrac{1}{6}x^3+\dfrac{3}{2}x^2+\dfrac{9}{2}x+\dfrac{9}{2} }}$$
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 $\bf{-3 }$ with a multiplicity of 3, so let's make it 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 $x=\bf{ -3 }$, $(x +3 )=0$
$(x+3)$ with a multiplicity 3 equals $(x+3)^3$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x+3 )(x+3 )(x+3 )}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x+3 )^3$
If $f( 3 )= 36 $ then
${( 3+3 )^3}$ times some number $a$, equals $ 36 $ or:
$f( 3 )= 36 =a(6)^3 $
$36 =a(6)^3$
$36=a(36)(6)$
$1=a(6)$
$\bf{a=\dfrac{1}{6} }$
Now that we have the value of $a$, we can find $f(x)$
$f(x)= \dfrac{1}{6}(x+3 )^3 $
$f(x)= \dfrac{1}{6}(x^2+6x+9)(x+3) $
$f(x)= \dfrac{1}{6}(x^3+6x^2+9x+3x^2+18x+27)$
$f(x)= \dfrac{1}{6}(x^3+9x^2+27x+27)$
$f(x)= \dfrac{1}{6}x^3+\dfrac{9}{6}x^2+\dfrac{27}{6}x+\dfrac{27}{6}$
$$\color{blue}{\bf{f(x)= \dfrac{1}{6}x^3+\dfrac{3}{2}x^2+\dfrac{9}{2}x+\dfrac{9}{2} }}$$
