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

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 ?)

• 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. Commented Aug 9, 2013 at 15:32
• @user64494 I don't have maple but can you please share the explanation that the software gave ? Commented Aug 9, 2013 at 15:49
• @ A Googler: See Maple worksheet exported as a PDF file here. Commented Aug 9, 2013 at 16:26
– user304329
Commented Nov 15, 2018 at 15:48

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

• 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 ? Commented Aug 9, 2013 at 10:53
• 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.
Commented Aug 9, 2013 at 11:15
• @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 $x$. Commented Aug 9, 2013 at 21:55
• @Eric ...unless you allow negative integers for $x$. The zero product allowed in that case is the only perfect square. Commented Feb 15, 2020 at 11:00

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.

• 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 ? Commented Aug 9, 2013 at 10:30
• @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. Commented Aug 9, 2013 at 10:44
• @AGoogler primitive polynomials with coefficients in Z are closed under multiplication. integer coefficients <=> all roots are in Z. Commented Aug 9, 2013 at 16:03

$$(x+1)(x+2)(x+3)(x+6)-3x²$$ $$(x+1)(x+6)(x+3)(x+2)-3x²$$ $$(x²+6x+1x+6)(x²+3x+2x+6)-3x²$$ $$(x²+6x+6+1x)(x²+5x+6+x-x)-3x²$$ $$(x²+6x+6+1x)(x²+5x+x+6-x)-3x²$$ $$(x²+6x+6+1x)(x²+6x+6-x)-3x²$$ $$(x²+6x+6)²(+1x)(-x)-3x²$$ $$(x²+6x+6)²-x²-3x²$$ $$(x²+6x+6)²-4x²$$ $$(x²+6x+6)²-(2x)²$$ 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: $$x²(x+8+6/x)(x+4+6/x)$$

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

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

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

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

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

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 :-

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

• here is a mathjax reference to typeset math. Commented Oct 27, 2020 at 4:52