1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ You need to say over which field you want the factorization, see Yves answer. $\endgroup$ Feb 3 '17 at 9:49
4
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ 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
2
$\begingroup$

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,

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.