Answer
$$\color{blue}{\bf{f(x)= -\dfrac{1}{2}x^3 +2x^2+\dfrac{11}{2}x-15 }}$$
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 $\bf{2 }$, $\bf{ -3 }$, and $\bf{5 }$, so let's make them into factors of our polynomial.
If $\bf{ 2 }$ is a zero then:
$( 2 +n)=0$
$n= -2 $
so our factor is $\bf{(x -2 )}$ because when $x=\bf{ 2 }$, $(x -2 )=0$
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 $\bf{ 5 }$ is a zero then:
$( 5 +n)=0$
$n= -5 $
so our factor is $\bf{(x-5 )}$ because when $x=\bf{ 5 }$, $(x -5 )=0$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x-2 )(x +3 )(x-5 )}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x-2 )(x +3 )(x-5 )$
If $f( 3 )= 6 $ then
${(3-2 )(3 +3 )(3-5 )}$ times some number $a$, equals $ 6 $ or:
$f( 3 )= 6 =a(3-2 )(3 +3 )(3-5 ) $
$6 =a(1 )(6 )(-2 ) $
$6 =a(-12)$
$-\dfrac{6}{12}=a$
$\bf{a= -\dfrac{1}{2} }$
Now that we have the value of $a$, we can find $f(x)$
$f(x)= -\dfrac{1}{2}(x-2 )(x +3 )(x-5 ) $
$f(x)= -\dfrac{1}{2}(x^2+x-6 )(x-5 ) $
$f(x)= -\dfrac{1}{2}(x^3 -4x^2-11x+30 ) $
$$\color{blue}{\bf{f(x)= -\dfrac{1}{2}x^3 +2x^2+\dfrac{11}{2}x-15 }}$$
