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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

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

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$$x^4 - 5x^3 - 5x^2 - 5x - 6=x^4-5(x^3+x^2+x+1)-1 \\=(x-1)(x^3+x^2+x+1)-5(x^3+x^2+x+1)$$

and $$x^3+x^2+x+1=(x+1)(x^2+1).$$

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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: $$P(x)=(x+1)(x-6)(x^2+1).$$

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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 $\Bbb Q[x]$.

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