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.

  • $\begingroup$ You need to say over which field you want the factorization, see Yves answer. $\endgroup$ Feb 3 '17 at 9:49

\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 $a=y^4,\ b=1$.

  • $\begingroup$ Note that this is not a general technique for factoring arbitrary polynomials, merely a trick that happens to work for this particular case. (Which may be all the OP wanted, of course -- it is probably all he can get). $\endgroup$ Feb 3 '17 at 9:24
  • $\begingroup$ $y^4+1$ can still be factored. $\endgroup$
    – user65203
    Feb 3 '17 at 9:32
  • $\begingroup$ Only in $\mathbb{C}$. $\endgroup$ Feb 3 '17 at 9:34
  • 1
    $\begingroup$ @RasmusErlemann: no, in $\mathbb R$. $\endgroup$
    – user65203
    Feb 3 '17 at 9:34
  • $\begingroup$ $y^4+1$ has no real roots. I don't know what you are talking about. $\endgroup$ Feb 3 '17 at 9:36

Factorization of $4y$ is trivial, leaving you with the polynomial


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 ($k=0,4$),


two are pure imaginary (conjugate, $k=2,6$),


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:



$$4y^9-4y=4y(y-1)(y+1)(y^2+1)(y^2-\sqrt2 y+1)(y^2+\sqrt2 y+1).$$

  • $\begingroup$ That is super complex for me, I hope I will learn that in future. $\endgroup$
    – Reboot
    Feb 3 '17 at 9:37
  • $\begingroup$ @Reboot: yep but leads to a more correct answer than the one you accepted. $y^4+1$ can be factored. $\endgroup$
    – user65203
    Feb 3 '17 at 9:39

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