How to factorize $x^5 -1=0$?
Matthew Harrington 
how can I factorize $x^5 -1 =0 $ so I get a quartic equation separately. Is there any specific way of factorizing such equations to get something of a lesser power?
$\endgroup$ 25 Answers
$\begingroup$Useful identity: $x^n-1=(x^{n-1}+\cdots+x+1)(x-1)$.
$\endgroup$ 4 $\begingroup$Divide $x^5-1$ by $x-1$ to get $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$ Now note that $$x^4+x^3+x^2+x+1=\left(x^2+\frac 12x+1\right)^2-\frac 54x^2$$ $$=\left(x^2+\frac 12x+1-\frac{\sqrt 5}{2}x\right)\left(x^2+\frac 12x+1+\frac{\sqrt 5}{2}x\right).$$
Hence, one has $$x^5-1=(x-1)\left(x^2+\frac{1-\sqrt 5}{2}x+1\right)\left(x^2+\frac{1+\sqrt 5}{2}x+1\right).$$
$\endgroup$ 1 $\begingroup$$$x^5-1=0\implies x^5=1$$
The solutions are the fifth roots of unity.
$$\large x=e^{2\pi i\frac{k}{5}}~,~k\in\Bbb{Z}~\land~0\leq k\lt 5$$
For each solution $a$, you have a factor $(x-a)$ thus giving you the five factors.
$\endgroup$ $\begingroup$The equation $z^n=\omega_0$ has the solution's:
$$z_k=\sqrt[n]{|\omega_0|}\left(cos\frac{\arg(\omega_0)+2k\pi}{n}\right)+i\sin\left(\frac{\arg(\omega_0)+2k\pi}{n}\right)\,\,,\,\,k=0,1,...,n-1$$
And the solution of the equation $z^n=1$, are $z_k=\cos(\frac{2k\pi}{n})+i\sin(\frac{2k\pi}{n})\,\,,\,\,k=0,1,...,n-1$
$\endgroup$ $\begingroup$Look for roots of the polynomial. For example, $x=1$ should be immediately apparent as a root of $x^5-1$. This means the quantity $x^5-1$ is divisible by $x-1$. Simply compute the polynomial division of $$\frac{x^5-1}{x-1} = P_4(x)$$ to get a fourth degree polynomial $P_4(x)$. Thus $x^5-1$ can be factored as $$x^5-1 = (x-1)P_4(x)$$ You could continue factoring $P_4(x)$, although the roots may be complex. If there are complex roots there will either be two or four of them.
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