Answer
$$\color{blue}{\bf{f(x)= -2x^3-8x^2 }}$$
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 $\bf{ -4}$, and $\bf{0 }$ of multiplicity 2, so let's make them into factors of our polynomial.
If $\bf{ -4 }$ is a zero then:
$( -4 +n)=0$
$n= 4 $
so our factor is $\bf{(x +4 )}$ because when $x=\bf{ -4 }$, $(x+4 )=0$
If $\bf{ 0 }$ is a zero then:
$( 0 +n)=0$
$n= 0 $
so our factor is $\bf{(x )}$ because when $x=\bf{ 0 }$, $(x +0 )=0$
Since $0$ is a zero of multiplicity 2, $(x)$ is a factor twice:
$(x)^2$ or $(x)(x)$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x +4 )(x )(x )}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x +4 )(x )(x )$
If $f( -1 )= -6 $ then
${(-1+4 )( -1 )( -1 )}$ times some number $a$, equals $ -6 $ or:
$f( -1 )= -6 =a(-1+4 )( -1 )( -1 ) $
$ -6 =a(3 )( -1 )( -1 ) $
$ -6 =a(3 ) $
$-\dfrac{6}{3}=a$
$\bf{a=-2 }$
Now that we have the value of $a$, we can find $f(x)$
$f(x)=-2 (x +4 )(x )(x ) $
$f(x)=-2 (x +4 )(x^2) $
$f(x)=-2 (x^3+4x^2) $
$$\color{blue}{\bf{f(x)= -2x^3-8x^2 }}$$