Factorise of polynomial

Factorize

$$x^4 - 5x^3 - 5x^2 - 5x - 6$$

I have tried different methods to solve but could no be able to do so. Please can somebody help. Your individual contributions would be greatly appreciated. Regards

• Inspection could be a good idea. – Claude Leibovici Apr 24 at 8:28
• Factor over which ring or field? – Servaes Apr 24 at 8:29
• Hint: $6 =5+1$. – Yves Daoust Apr 24 at 9:00

$$x^4 - 5x^3 - 5x^2 - 5x - 6=x^4-5(x^3+x^2+x+1)-1 \\=(x-1)(x^3+x^2+x+1)-5(x^3+x^2+x+1)$$
and $$x^3+x^2+x+1=(x+1)(x^2+1).$$
You can easily check for some solutions and then factorize the polynomial. For instance, if you set $$x=-1$$, then $$P(-1)=1+5-5+5-6=0$$, so $$x=-1$$ is one of the roots.
Now, you can divide the polynomial to $$x+1$$. We have: $$\frac{P(x)}{x+1}=x^3-6x^2+x-6$$ and you can easily see that: $$x^3-6x^2+x-6=(x-6)(x^2+1)$$ So: $$P(x)=(x+1)(x-6)(x^2+1).$$
Obvioulsy, $$x=-1$$ and $$x=6$$ are roots, so that we obtain $$x^4 - 5x^3 - 5x^2 - 5x - 6=(x^2+1)(x+1)(x-6)$$ in $$\Bbb Q[x]$$.