Answer
$f(x)=2x^{3}-20x^{2}+64x-64$
Work Step by Step
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
So,
$(x-2)$ is a factor of f,
$(x-4)$ 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,
$(x-2)$ is a factor of f, degree=1
$(x-4)^{2}$ is a factor of f, degree = 2
(the sum of degrees is 3)
Since f has degree 3, $\quad f(x)=a(x-2)(x-4)^{2}$
To find $a,$ use the given information: $f(1)=-18$
$a(1-2)(1-4)^{2}=-18$
$a(-1)(9)=-18$
$-9a=-18$
$a=2$
Thus,
$f(x)=2(x-2)(x-4)^{2}$
$=(2x-4)(x^{2}-8x+16 )$
$=2x\left(x^{2}-8x+16\right)-4\left(x^{2}-8x+16\right)$
$=2x^{3}-16x^{2}+32x-4x^{2}+32x-64$
$f(x)=2x^{3}-20x^{2}+64x-64$