Question 1029942
Zeros are  2 and 4 and 4 again.


{{{f(x)=k(x-2)(x-4)^2}}}, which takes care of the zeros.
{{{f(x)=k(x-2)(x^2-8x+16)}}}
{{{k(x^3-8x^2+16x-2x^2+16x-32)}}}
{{{highlight(k(x^3-10x^2+32x-32))}}}, and then seems your choice to multiply through by k or not.


Finding k should be easier using the fully factored form.
{{{f(1)=k(1-2)(1-4)^2=-18}}}
{{{k(-1)(-3)^2=-18}}}
{{{k(-1)*9=-18}}}
{{{k=2}}}


Now you can multiply through using k=2.
{{{highlight(highlight(f(x)=2x^3-20x^2+64x-64))}}}