AoPS Wiki talk:Problem of the Day/July 1, 2011


AoPSWiki:Problem of the Day/July 1, 2011

Solution 1

To factor $(n+1)(n+2)(n+3)(n+4) - 120$, we should try to find a way to create a quadratic in disguise. There, in fact, is a way!

Expand $(n+1)(n+4)$ and $(n+2)(n+3)$ separately:

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

We notice that there is a $n^2 + 5n$ in both of these terms! Treat $n^2 + 5n$ as a single quantity, $x$.

Our original quantity was $(x + 4)(x+6)-120 = x^2+10x+24-120=x^2+10x-96=(x-6)(x+16)$.

We can factor the original polynomial as $[(n^2+5n)-6][(n^2+5n)+16] = (n^2+5n-6)(n^2+5n+16)$.

$n^2+5n-6$ can be factored as $(n-1)(n+6)$.

Our final factorization is $\boxed{(n-1)(n+6)(n^2+5n+16)}$.

Solution 2

Let $P(x) = (x+1)(x+2)(x+3)(x+4) - 120$. Then $P(1) = 0$ and $P(-6)=0$ (since $120 = -5\cdot -4\cdot -3\cdot -2$). Therefore I have: \[P(x) = (x-1)(x+6)Q(x)\] where $Q$ is a quadratic. Simplifying yields $Q(x)=x^2+5x+16$ as before, which is irreducible as $16 > \frac{5^2}{4}$.