Answer
$$\color{blue}{\bf{f(x)= 2x ^3-20x^2+64x-64 }}$$
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 $\bf{ 2 }$, and $\bf{4 }$ with a multiplicity of 2 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 $\bf{ 4 }$ is a zero then:
$( 4 +n)=0$
$n= -4 $
so our factor is ${(x -4 )}$ because when $x=\bf{ 4 }$, $(x -4 )=0$
Since $4$ is a zero with multiplicity 2, $(x-4)$ is a factor twice:
$\bf{(x-4)^2}$ or $\bf{(x-4)(x-4)}$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x -2 )(x-4 )(x -4 )}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x -2 )(x -4 )^2$
If $f( 1 )= -18 $ then
${(1 -2 )(1 -4 )^2}$ times some number $a$, equals $ -18 $ or:
$f( 1 )= -18 =a(1 -2 )(1 -4 )^2 $
$ -18 =a(-1 )(-3 )^2 $
$ -18 =a(-1 )(9) $
$-18 =a(-9) $
$ \dfrac{-18}{-9} =a \dfrac{-9}{-9} $
$\bf{a= 2 }$
Now that we have the value of $a$, we can find $f(x)$
$f(x)= 2(x -2 )(x -4 )^2 $
$f(x)= 2(x -2 )(x -4 ) (x -4 ) $
$f(x)= 2(x -2 )(x^2-8x+16) $
$f(x)=2(x^3-2x^2-8x^2+16x+16x-32)$
$f(x)= 2(x ^3-10x^2+32x-32) $
$$\color{blue}{\bf{f(x)= 2x ^3-20x^2+64x-64 }}$$
