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 $\mathbb{Z}$.
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1Have you tried factorising it into quadratics? – user88595 May 30 '14 at 16:11
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how can i find such quadratics – cf12418 May 30 '14 at 16:13
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3try $(x^2 + A x + B)(x^2 + C x + D)$ – Will Jagy May 30 '14 at 16:14
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surely i need a root to do this? there will be too many constants? – cf12418 May 30 '14 at 16:14
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1$x^4 + 3 x^2 + 6 x + 10 = \left(x^2-2 x+5\right) \left(x^2+2 x+2\right)$ – bonanza May 30 '14 at 16:15
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2There 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. – Hans Engler May 30 '14 at 16:16
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4@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. – Git Gud May 30 '14 at 16:16
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Related posts, with answers: 1 2 – MJD May 30 '14 at 16:23
1 Answers
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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 $\pm10$.
user88595
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