6
$\begingroup$

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 $\mathbb{Z}$.

$\endgroup$
8
  • 1
    $\begingroup$ Have you tried factorising it into quadratics? $\endgroup$
    – user88595
    May 30 '14 at 16:11
  • 3
    $\begingroup$ try $(x^2 + A x + B)(x^2 + C x + D)$ $\endgroup$
    – Will Jagy
    May 30 '14 at 16:14
  • 1
    $\begingroup$ $x^4 + 3 x^2 + 6 x + 10 = \left(x^2-2 x+5\right) \left(x^2+2 x+2\right)$ $\endgroup$
    – bonanza
    May 30 '14 at 16:15
  • 2
    $\begingroup$ 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. $\endgroup$ May 30 '14 at 16:16
  • 4
    $\begingroup$ @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. $\endgroup$
    – Git Gud
    May 30 '14 at 16:16
8
$\begingroup$

Let : $$x^4 + 3x^2 + 6x + 10 = (x^2 + ax + b)(x^2 + cx + d)$$

Expanding the right hand side and matching coefficients with the left hand side you need to solve the following equations : \begin{eqnarray*} a + c &=& 0 \qquad (x^3)\\ b + d + ac &=& 3 \qquad (x^2)\\ ad + bc &=& 6 \qquad (x)\\ bd &=& 10 \qquad (x^0) \end{eqnarray*}

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 $\pm10$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.