Factoring $(x+1)(x+2)(x+3)(x+4)+1$

Question

I have to factor this polynomial: $$(x+1)(x+2)(x+3)(x+4)+1$$ WolframAlpha gives $$(x^2+5x+5)^2$$ How can I prove it without expanding the result?

Thanks!

If you can pretend you "guessed" the factorization, notice $(x^2+5x+5)^2-1$ is a difference of squares. — Jean-Claude Arbaut, Aug 25 '14 at 14:21
Use $x^2 + 5x + 6 = (x+2)(x+3)$ and $x^2 + 5x + 4 = (x+1)(x+4)$ and Jean-Claude's tip above. — Winther, Aug 25 '14 at 14:23
@gammatester but then you will only prove that the set of roots from one polynomial is included in the set of roots of the other. Stay to show that $(x+1)(x+2)(x+3)(x+4)+1$ has two distinct roots of multiplicity $2$... — Surb, Aug 25 '14 at 14:24

score 13 · Answer 1 · answered Aug 25 '14 at 14:21

13

Evaluate both polynomials at $x = -1, -2, -3,$ and $-4$. They are equal at these four values, and their leading coefficients are both equal to $1$. So they are the same polynomial.

answered Aug 25 '14 at 14:21

TonyK

64,559

Simply beautiful! – Surb Aug 25 '14 at 14:22
So simple when you see it, nice! – Jean-Claude Arbaut Aug 25 '14 at 14:22
I was just about writing the same solution. To clarify: 4 points fixate a polygon of degree 4. – Jakube Aug 25 '14 at 14:23
@TonyK Thanks! What if the result was unknown? – e-man Aug 25 '14 at 14:23
1

@e-man: Ah, now that's a different question :-) – TonyK Aug 25 '14 at 14:28

score 10 · Accepted Answer · edited Apr 13 '17 at 12:21

10

I answered a similar question here. You can do it like this: $(x+2)(x+3) = x^2+5x+5+1$, $(x+1)(x+4) = x^2+5x+5-1$, so $$ (x+1)(x+2)(x+3)(x+4) = (x^2+5x+5+1)(x^2+5x+5-1) = (x^2+5x+5)^2-1^2. $$ This gives the answer $(x+1)(x+2)(x+3)(x+4)+1 =(x^2+5x+5)^2 $.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 25 '14 at 14:26

N.U.

3,103

Wow, I like it! – e-man Aug 25 '14 at 14:29

score 5 · Answer 3 · answered Aug 25 '14 at 14:28

5

$(x^2+5x+5)^2-1^2=(x^2+5x+4)(x^2+5x+6)=(x+1)(x+4)(x+2)(x+3)$

answered Aug 25 '14 at 14:28

drhab

151,093

Your answer is a clear one! +1 – Olivier Oloa Aug 25 '14 at 17:22

score 3 · Answer 4 · answered Aug 25 '14 at 14:37

The key for this problem is observing that $(x+1)(x+2)(x+3)(x+4)+1$ is symmetric with respect to $x=-2.5$. Write it as

$$(x+1)(x+2)(x+3)(x+4)+1= \\ \left( (x+2.5)-1.5\right)\left( (x+2.5)-0.5\right)\left((x+2.5)+0.5 \right)\left((x+2.5)+1.5 \right)+1$$

For simplicity, let us denote temporarily $x+2.5$ by $y$. Then we need to factor

$$\left( y-1.5\right)\left(y+1.5 \right)\left(y-0.5\right)\left(y+0.5 \right)+1=\left( y^2-1.5^2\right)\left(y^2-0.5^2\right)+1$$

Now, this is a quadratic in $y^2$, which can be factored by completing the square. Then replace $y=x+2.5$.

Bill Dubuque · Answer 5 · 2014-08-25T15:34:00.910

Lemma $\ ad\!-\!bc = -2\,\Rightarrow\, abcd = e^2\!-1,\,\ e = ad\!+\!1 = bc\!-\!1$

Proof $\ $ Notice that $\ abcd = (ad)(bc) = (e\!-\!1)(e\!+\!1) = e^2\!-1\ $ QED

Yours is the special case $\ a,b,c,d\, =\, x\!+\!1,x\!+\!2,x\!+\!3,x\!+\!4$

Generally $\ \ AB\, =\, \left(\dfrac{A+B}2\right)^2-\left(\dfrac{A-B}2\right)^2 $

The Lemma is the special case $\,\ |A\!-\!B|\, =\, 2.\ $ This identity often proves useful when one wishes to replaces multiplication by squaring(s) - just as in your problem.

Factoring $(x+1)(x+2)(x+3)(x+4)+1$

5 Answers5