# Factorise $x^4 + 3x^2+ 6x+ 10$

I need to factorise $x^4 + 3x^2 + 6x + 10$ completely over $\mathbb{Q}$.

I am not sure how to do this. I can't find any roots of this equation in $\mathbb{Z}$.

• Have you tried factorising it into quadratics? May 30 '14 at 16:11
• try $(x^2 + A x + B)(x^2 + C x + D)$ May 30 '14 at 16:14
• $x^4 + 3 x^2 + 6 x + 10 = \left(x^2-2 x+5\right) \left(x^2+2 x+2\right)$ May 30 '14 at 16:15
• There are four constants $A, B, C, D$ and you know four coefficients of the result. Also, $BD = 10$, so there are only a few possible choices. May 30 '14 at 16:16
• @cf12418 Use Gauss's lemma. You can factorize it over $\mathbb Q$ if, and only if, you can do it over $\mathbb Z$. Thus Will's hint suffices. May 30 '14 at 16:16

Let : $$x^4 + 3x^2 + 6x + 10 = (x^2 + ax + b)(x^2 + cx + d)$$
Clearly $a = -c$ and then you could try $b = \pm1, d = \pm10$ or $b = \pm2, d = \pm5$ and see what you get for $a$ and $c$.
Something else you can notice from the third equation is that $a(d-b) = 6$. This leads to the fact that $d-b|6$ hence they can't be $\pm1$ and $\pm10$.