Answer
$f(x)=\displaystyle \frac{1}{6}x^{3}+\frac{3}{2}x^{2}+\frac{9}{2}x+\frac{9}{2}$
Work Step by Step
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
So, $(x+3)$ is a factor of f.
The number of times $(x-k)$ occurs as a factor is referred to as the multiplicity of the zero.
So, $(x+3)^{3}$ is a factor of f.
Since f has degree 3, $\quad f(x)=a(x+3)^{3}.$
To find $a,$ use the given information: $f(3)=36$
$a(3+3)^{3}=36.$
$a\cdot 6^{3}=36$
$a=\displaystyle \frac{6^{2}}{6^{3}}=\frac{1}{6}$
Thus,
$f(x)=\displaystyle \frac{1}{6}(x+3)^{3}$
$=\displaystyle \frac{1}{6}\left(x^{3}+3\cdot x^{2}\cdot 3+3\cdot x\cdot 3^{2}+3^{3}\right)$
$=\displaystyle \frac{1}{6}\left(x^{3}+9x^{2}+27x+27\right)$
$=\displaystyle \frac{1}{6}x^{3}+9\cdot\frac{1}{6}x^{2}+27\cdot\frac{1}{6}x+27\cdot\frac{1}{6}$
$f(x)=\displaystyle \frac{1}{6}x^{3}+\frac{3}{2}x^{2}+\frac{9}{2}x+\frac{9}{2}$
