0
$\begingroup$

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.

  • 0
    You need to say over which field you want the factorization, see Yves answer. – 2017-02-03

2 Answers 2

4

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

  • 0
    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_). – 2017-02-03
  • 0
    $y^4+1$ can still be factored. – 2017-02-03
  • 0
    Only in $\mathbb{C}$. – 2017-02-03
  • 1
    @RasmusErlemann: no, in $\mathbb R$. – 2017-02-03
  • 0
    $y^4+1$ has no real roots. I don't know what you are talking about. – 2017-02-03
  • 0
    If $y^4 + 1$ would not factor in $\mathbb{R}$, $\mathbb{C}$ wouldn't be a quadratic extension of $\mathbb{R}$. – 2017-02-03
  • 0
    I am talking about $(y^2-\sqrt2y+1)(y^2+\sqrt2y+1)$. – 2017-02-03
  • 0
    I think OP was looking for factorization by the real roots of the polynomial. Overall you are right, interesting approach. – 2017-02-03
  • 0
    @RasmusErlemann: more precisely, you performed the factorization by the real roots, believing that this is enough; the OP didn't say anything. Then for him the answer is globally "no, there is no easy way". – 2017-02-03
2

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,

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

  • 0
    That is super complex for me, I hope I will learn that in future. – 2017-02-03
  • 0
    @Reboot: yep but leads to a more correct answer than the one you accepted. $y^4+1$ can be factored. – 2017-02-03