Is there easy way to factor polynomials for calculus? factor of $$4y^9-4y$$
comes out to be $$4y(y^4+1)(y^2+1)(y+1)(y-1)$$
How would you approach to factor?
PS: My math is very rusty.
 A: \begin{align*}
4y^9-4y&=4y(y^8-1)\\
&=4y((y^4)^2-1)\\
&=4y(y^4-1)(y^4+1)\\
&=4y((y^2)^2-1)(y^4+1)\\
&=4y(y^2-1)(y^2+1)(y^4+1)\\
&=4y(y-1)(y+1)(y^2+1)(y^4+1).
\end{align*}
I am simply using $a^2-b^2=(a-b)(a+b)$. For example, at the second equation we took $a=y^4,\ b=1$.
A: Factorization of $4y$ is trivial, leaving you with the polynomial
$$y^8-1$$
Then the factorization of any polynomial is the product of the monomials $(y-r_k)$ where $r_k$ are the roots, possibly taken with their multiplicites. So you need to solve
$$y^8=1,$$
or
$$"y=\sqrt[8]1".$$
The computation of the eighth roots must be made in the complex and can be carried out using the polar form:
$$y^8=e^{i2k\pi}\iff y=e^{ik\pi/4}.$$
Among the solutions, two are real ($k=0,4$),
$$y=1,y=-1$$
two are pure imaginary (conjugate, $k=2,6$),
$$y=i,y=-i$$
and the four remaining ones are complex (pairwise conjugate, $k=1,3,5,7$)
$$\pm\frac1{\sqrt2}\pm\frac i{\sqrt2}.$$
If you don't want the imaginary/complex numbers to appear, you can keep the quadratic monomials resulting from conjugate pairs:
$$(y-a-ib)(y-a+ib)=y^2-2ay+a^2+b^2.$$
Finally,
$$4y^9-4y=4y(y-1)(y+1)(y^2+1)(y^2-\sqrt2 y+1)(y^2+\sqrt2 y+1).$$
