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I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$ I don't really know where to start, but I expanded everything to get : $$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what should I do to factorize this polynomial ?

(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)

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The Maple command $$ infolevel[factor] := 5: factor((x+1)*(x+2)*(x+3)*(x+6)-3*x^2)$$ produces $$(x^2+8*x+6)*(x^2+4*x+6) $$ and the explanation how this is found. –  user64494 Aug 9 '13 at 15:32
    
@user64494 I don't have maple but can you please share the explanation that the software gave ? –  A Googler Aug 9 '13 at 15:49
    
@ A Googler: See Maple worksheet exported as a PDF file here. –  user64494 Aug 9 '13 at 16:26

2 Answers 2

up vote 38 down vote accepted

A way to do it is to write $(x+2)(x+3) = x^2 + 6x + 6 - x$, $(x+1)(x+6) = x^2 + 6x + 6 + x$, so

$$ (x+2)(x+3)(x+1)(x+6) = (x^2 + 6x + 6)^2 -x^2 $$ which gives that $$ (x+2)(x+3)(x+1)(x+6) - 3x^2 = (x^2 + 6x + 6)^2 -4x^2 = (x^2+ 4x + 6)(x^2 +8x + 6). $$

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Wonderful ! But I'm curious to know how did you realize to split (x+2)(x+3)(x+1)(x+6) into two parts (x+2)(x+3) and (x+1)(x+6) ? Also is this because 2*3=1*6=6 ? Does this method always work if in {(x+a)(x+b)(x+c)(x+d)} , ab=cd ? –  A Googler Aug 9 '13 at 10:53
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Yes it was because 2*3=1*6 = 6 and that the whole expression was nearly on the form $x^2-y^2$, so this is not a very general method and does not work in general cases. –  N.U. Aug 9 '13 at 11:15
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@AGoogler: Experience. Such groupings often come up. For example, prove that $$(x+1)(x+2)(x+3)(x+4)$$ is never a perfect square for any $x$. –  Eric Naslund Aug 9 '13 at 21:55

I'm assuming that you are looking for a factorization of the polynomial $$ f = x^4 + 12x^3 + 44x^2 + 72x + 36 $$ in $\mathbb Q[x]$. By the Lemma of Gauss and the fact that $f$ is monic, this is the same as looking for a factorization in $\mathbb Z[x]$.

Since there are no rational roots, the only remaining possibility is the factorization into two factors of degree 2. Since the polynomial is monic, the factors may be assumed to be monic. So the putative factors have the form $x^2 + ax + b$ and $x^2 + cx + d$ with $a,b,c,d\in\mathbb Z$.

This leads to $$ f = (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd. $$ Comparing the coefficients, we get four equations $$ a+c = 12\\ ac + b + d = 44\\ ad + bc = 72\\ bd = 36 $$ Up to swapping the two factors, the last equation has the solutions $$ (b,d)\in\{(1,36),(-1,-36),(2,18),(-2,-18),(3,12),(-3,-12),(4,9),(-4,-9),(6,6),(-6,-6)\}. $$ It's a bit of work, but plugging these values into the remaining three equations, you find that only for $b=d=6$ there is a solution, which is $a = 4$, $b = 8$ or $a = 8$, $b=4$. This gives you the two factors.

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Thanks for the answer ! I searched for Gauss's lemma and found a Wikipedia article . But I didn't really understood it . Can you please explain the lemma ? –  A Googler Aug 9 '13 at 10:30
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@AGoogler: I applied the corollary from Grauss's lemma that a primitive polynomial with integer coefficients is irreducible over the integers iff it is irreducible over the rationals. –  azimut Aug 9 '13 at 10:44
    
@AGoogler primitive polynomials with coefficients in Z are closed under multiplication. integer coefficients <=> all roots are in Z. –  glebm Aug 9 '13 at 16:03

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