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