<s>[INST] How to factorize $x^5 -1=0$? 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?
 [/INST] 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
$$x^5-1=(x-1)\left(x^2+\frac{1-\sqrt 5}{2}x+1\right)\left(x^2+\frac{1+\sqrt 5}{2}x+1\right).$$
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