The polynomial to be factorised as a product of two factors is- $$x^4+3x^2+6x+10$$. I checked the solution in wolfram alpha to be- $$(x^2-2x+5)(x^2+2x+2)$$. I tried to factorise it by expressing it as a sum of two squares $$(x^2+1)^2+(x+3)^2$$. But I cannot solve it. Please help. Thanks a lot in advance.

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  • $\begingroup$ I tried using factor theorem also. $\endgroup$ – tatan Sep 11 '15 at 9:18
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    $\begingroup$ since you know that $x^3$ terms vanish we can assume that the linear terms for $x$ are equal and opposite. $\endgroup$ – Chinny84 Sep 11 '15 at 9:27
  • $\begingroup$ @Chinny84-Please give a step by step solution. $\endgroup$ – tatan Sep 11 '15 at 9:28
  • $\begingroup$ If you want the full factorization in $\Bbb C$, your attempt is not so bad. You can write $$x^4+3x^2+6x+10=(x^2+1)^2+(x+3)^2=(x^2+1)^2-(ix+3i)^2=(x^2+ix+1+3i)(x^2-ix+1-3i)$$ Then factor the trinomials, and you will get the $4$ complex roots (there is obviously no real root). $\endgroup$ – Jean-Claude Arbaut Jun 7 '18 at 9:20

This is going beyond my knowledge in irreducible/reducible forms.. but it seems clear that if we assume at first glance that we have (which is not the way I would go about factoring forms like this) $$ \left(x^2+ax\cdots\right)\left(x^2+bx\cdots\right) $$ then we know that $$ \left(x^2+ax\cdots\right)\left(x^2+bx\cdots\right) = x^4 +(a+b)x^3 + \cdots $$ thus we must have $a+b = 0\implies a = -b$

but like I said I know there must be some group theory approach thanks to Galois et al.

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  • $\begingroup$ Nice intuition!!! $\endgroup$ – tatan Sep 11 '15 at 9:42
  • $\begingroup$ Thanks. Though I believe there exists a nicer solution out there :)!! $\endgroup$ – Chinny84 Sep 11 '15 at 9:44

The given polynomial $x^4+3x^2+6x+10$ Can be written as

$ x^4+2x^3+2x^2-2x^3-4x^2-4x+5x^2+10x+10$

$= x^2(x^2+2x+2) - 2x(x^2+2x+2) +5(x^2+2x+2)$

$=(x^2+2x+2)(x^2-2x+5) $

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