33
$\begingroup$

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 ?)

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – user64494
    Aug 9, 2013 at 15:32
  • $\begingroup$ @user64494 I don't have maple but can you please share the explanation that the software gave ? $\endgroup$
    – A Googler
    Aug 9, 2013 at 15:49
  • $\begingroup$ @ A Googler: See Maple worksheet exported as a PDF file here. $\endgroup$
    – user64494
    Aug 9, 2013 at 16:26
  • 1
    $\begingroup$ Don't post PDF files as answers. Type them out so future users can still read it, even if your link dies. $\endgroup$
    – user304329
    Nov 15, 2018 at 15:48

4 Answers 4

Reset to default
46
$\begingroup$

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

$\endgroup$
4
  • 7
    $\begingroup$ 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 ? $\endgroup$
    – A Googler
    Aug 9, 2013 at 10:53
  • 6
    $\begingroup$ 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. $\endgroup$
    – N.U.
    Aug 9, 2013 at 11:15
  • 4
    $\begingroup$ @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 $x$. $\endgroup$
    – Eric Naslund
    Aug 9, 2013 at 21:55
  • $\begingroup$ @Eric ...unless you allow negative integers for $x$. The zero product allowed in that case is the only perfect square. $\endgroup$
    – Oscar Lanzi
    Feb 15, 2020 at 11:00
33
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 ? $\endgroup$
    – A Googler
    Aug 9, 2013 at 10:30
  • 1
    $\begingroup$ @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. $\endgroup$
    – azimut
    Aug 9, 2013 at 10:44
  • $\begingroup$ @AGoogler primitive polynomials with coefficients in Z are closed under multiplication. integer coefficients <=> all roots are in Z. $\endgroup$
    – glebm
    Aug 9, 2013 at 16:03
2
$\begingroup$

$$(x+1)(x+2)(x+3)(x+6)-3x²$$ $$(x+1)(x+6)(x+3)(x+2)-3x²$$ $$(x²+6x+1x+6)(x²+3x+2x+6)-3x²$$ $$(x²+6x+6+1x)(x²+5x+6+x-x)-3x²$$ $$(x²+6x+6+1x)(x²+5x+x+6-x)-3x²$$ $$(x²+6x+6+1x)(x²+6x+6-x)-3x²$$ $$(x²+6x+6)²(+1x)(-x)-3x²$$ $$(x²+6x+6)²-x²-3x²$$ $$(x²+6x+6)²-4x²$$ $$(x²+6x+6)²-(2x)²$$ Apply $a²-b²=(a+b)(a-b)$ $$(x²+6x+6+2x)(x²+6x+6-2x)$$ $$(x²+8x+6)(x²+6x-2x+6)$$ $$(x²+8x+6)(x²+4x+6)$$ $$x(x+8+6/x)x(x+4+6/x)$$ Answer: $$x²(x+8+6/x)(x+4+6/x)$$

$\endgroup$
2
$\begingroup$

$(x+1)(x+2)(x+3)(x+6)-3x^2$

$\rightarrow(x+1)(x+6)(x+2)(x+3)-3x^2$

$\rightarrow(x^2+x+6x+6)(x^2+2x+3x+6)-3x^2$

$\rightarrow(x^2+7x+6)(x^2+5x+6)-3x^2$

$\rightarrow(x^2+6+7x)(x^2+6+5x)-3x^2$

Put :- $x^2+6=y$ eq($1$) . So :-

$(y+7x)(y+5x)-3x^2$

$\rightarrow y^2+7xy+5xy+35x^2-3x^2$

$\rightarrow y^2+12xy+32x^2$

$\rightarrow y^2+4xy+8xy+32x^2$

$\rightarrow y(y+4x)+8x(y+4x)$

$\rightarrow (y+4x)(y+8x)$

Put value of $y$ from eq(1), we get :-

$\rightarrow (x^2+6+4x)(x^2+6+8x)$

$\endgroup$
1
  • $\begingroup$ here is a mathjax reference to typeset math. $\endgroup$
    – Siong Thye Goh
    Oct 27, 2020 at 4:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .