Not understanding how to factor a polynomial completely $$P(x)=16x^4-81$$
I know that this factors out as:
$$P(x)=16(x-\frac { 3 }{ 2 } )^4$$
What I don't understand is the four different zeros of the polynomial...I see one zero which is $\frac { 3 }{ 2 }$ but not the three others.

You have to solve the expression $16x^4 - 81 = 0$, and you will get
$$ 16x^4 - 81 = (4x^2 - 9)(4x^2 + 9) = 0 $$
then you will find the other three roots you didn't find.
