Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?
 A: Or, alternatively, note that $$x^4-x^3+5x^2-4x+4=x^2(x^2-x+1)+4x^2-4x+4$$ and factor $4$ from the last three terms.
A: Here I think this will work . Main idea being splitting up of $5x^2$ as $4x^2 + x^2$
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
$$(x+1) (x^4-x^3+4x^2+ x^2-4x+4)=0 $$
$$(x+1) (x^4+4x^2+ x^2-x^3-4x+4)=0 $$
$$(x+1)[ x^2(x^2+4)+ (1-x)( x^2 +4)]=0 $$
$$(x+1)[ (x^2+4)(x^2+1-x))=0 $$
$$(x+1)[ (x^2+4)(x^2-x+1))=0 $$
Hence proved 
A: I would start by factoring $x^3$ out from the first two terms and noticing the pattern in the result.
$$
\begin{split}
x^5+4x^3 + x^2 + 4 &= x^3 \left(x^2+4\right) + x^2+4 \\
                   &= \left(x^2+4\right)\left(x^3+1\right) \\
                   &= \left(x^2+4\right)(x+1)\left(x^2-x+1\right), \\
\end{split}
$$
where the last step is the standard factoring of the sum of two cubes.
A: The answers actually
$$
(x^2 + 4)(x^3 + 1)\\
x^5 + 4x^3 + x^2 + 4\\
(x^5 + 4x^3)(x^2 + 4)\\
x^3(x^2 + 4) 1(x^2 + 4)\\
(x^2 + 4)(x^3 + 1)
$$
(Sorry, too lazy to actually change them to exponents)
