Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$

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.

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.

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