Question 253571
We have 2 zeros at 2 and 2.
The third zero is 
{{{1 - isqrt(3)}}}
The fourth zero is the conjugat of this which is
{{{1 + isqrt(3)}}}
Remember: imaginaries always occur in pairs.
So, we have
{{{P(x) = (x-2)(x-2)(x - (1 - isqrt(3)))(x - (1 + isqrt(3)))}}}
The product of these factors is
{{{P(x) = (x^2 -4x + 4)(x^2 -2x + 2)}}}and then
{{{P(x) = X^4 -10x^3 + 14x^2 -16x + 8}}}
But we wanted 32 as the constant, so multiply by 4 to get
{{{P(x) = 4X^4 -40x^3 + 56x^2 -64x + 32}}}