<s>[INST] How to factorize $P=(X^2-4X+1)^2+(3X-5)^2$ in $\mathbb R[X]$? First of all, I searched for roots. I knew that $\exists x\in\mathbb R,\, P(x)=0 \iff \exists x\in\mathbb R,\, x^2-4x+1 = 0 \text{ and } 3x-5 = 0$.
However, It is really easy to say that there is no real roots. So there is only complex non real roots.
And we know that $\deg{(P)}=4$ and $\text{dom}(P)=1$.
Which means that $P=(X^2+aX+b)(X^2+cX+d) = X^4+(a+c)X^3 + (ac+b+d) X^2 + (ad+bc) X+ bd $
With $a^2-4b<0$ , $c^2-4d<0$ and $a,b,c,d\in\mathbb R$ .
$P = (X^2-4X+1)^2+(3X-5)^2 = X^4 - 8 X^3 + 27 X^2 - 38 X + 26$. This gives us a system :
$$\left\{\begin{align}
a+c &= -8\\
ac+b+d&=27\\
ad+bc&=-38\\
bd&=26
\end{align}\right .
$$
However, I tried and i don't know how to solve this system.
I tried secondly with an euclidean division.
I found that $P = (X^2+aX+b)[X^2+(-8-a)X+(27-b+8a+a^2)]+(-38+8b+2ab-27a-8a^2-a^3)X+(-1+b^2-8ab-a^2b)$
With $a,b\in\mathbb R,\, a^2-4b <0$ and $(-8-a)^2-4(27-b+8a+a^2)<0$.
We have now this other system :
$$\left\{\begin{align}
-38+8b+2ab-27a-8a^2-a^3 &= 0\\
-1+b^2-8ab-a^2b&=0
\end{align}\right .
$$
However, I don't know how to solve this.
How would you do this ? And if it is the same thing as me, how could you solve those systems ?
 [/INST] Assume (hope) that $a,b,c,d\in\Bbb Z$. If $bd=26$ then either:

*

*$b=1,d=26$ or $b=26,d=1$ which gives $ac=0$ but this  is absurd;


*$b=2,d=13$ which gives $13a+2c=-38$, equating with $a+c=-8$ gives $a=-2,c=-6$.
Indeed $(X^2-4X+1)^2+(3X-5)^2=(X^2-2X+2)(X^2-6X+13)$. This is the unique irreducible factorisation so we are done.
</s>