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.

Try making the substitutions $a=2x$ and $b=3$.  Then we have that
$$
16x^4-81=a^4-b^4.
$$
Recalling that $x^2-y^2=(x+y)(x-y)$, we have
$$
a^4-b^4=[a^2-b^2](a^2+b^2)=[(a-b)(a+b)](a^2+b^2)
$$
Plugging $2x$ in for $a$ and $3$ in for $b$ we have
$$
(a-b)(a+b)(a^2+b^2)=(2x-3)(2x+3)((2x)^2+3^2)=(2x-3)(2x+3)(4x^2+9).
$$
If you set each of these three factors equal to zero you will find the four roots you are looking for.
