Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 1, 2011"

Revision as of 20:22, 30 June 2011

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 $[(n^2+5n)-6][(n^2+5n)+16] = (n^2+5n-6)(n^2+5n+16)$ .

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

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

@@ Line 12: / Line 12: @@
 Our original quantity was <math>(x + 4)(x+6)-120 = x^2+10x+24-120=x^2+10x-96=(x-6)(x+16)</math>.
-We can factor the original polynomial as <math>[(n^2+5n)-6][(n^2+5n)+16] = (n^2+5n-6)(n^2+5n+16)</math>
+We can factor the original polynomial as <math>[(n^2+5n)-6][(n^2+5n)+16] = (n^2+5n-6)(n^2+5n+16)</math>.
 Notice that <math>n^2+5n-6</math> can be factored as <math>(n-1)(n+6)</math>.
 Our final factorization is <math>\boxed{(n-1)(n+6)(n^2+5n+16)}</math>.