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.
 A: Recognize $16x^4-81$ as a difference of two squares.
Then factor into linear factors,
$$
\begin{align*}
P(x)&=16x^4-81\\
&=(4x^2)^2-9^2\\
&=(4x^2-9)(4x^2+9)\\
&=(2x-3)(2x+3)(2x+3i)(2x-3i).
\end{align*}
$$
Now $P(x)=0$ iff
$$
2x\pm3=0
\qquad\text{or}\qquad
2x\pm3i=0
$$
iff
$$
x=\pm\frac32
\qquad\text{or}\qquad
x=\pm\frac32i.
$$
So we've found all four roots.
A: 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.
A: The four zeros are $\frac{3}{2},\frac{-3}{2},\frac{3i}{2},\frac{-3i}{2}$
The $n$th root of a number has $n$ answers.
To find each value of the $n$th root of $x$, draw a unit circle. The principal root is simply horizontal. Take that line and rotate it about the origin by increments of $\dfrac{2\pi}{n}$ to get each value. The y axis is imaginary and the x axis is real.
A: 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.
