Factoring $(x+1)(x+2)(x+3)(x+4)+1$ 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!
 A: $(x^2+5x+5)^2-1^2=(x^2+5x+4)(x^2+5x+6)=(x+1)(x+4)(x+2)(x+3)$
A: 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 $y=x+2.5$. 
A: 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.
A: 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.
A: 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 $.
