Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 1, 2011"
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{{:AoPSWiki:Problem of the Day/July 1, 2011}} | {{:AoPSWiki:Problem of the Day/July 1, 2011}} | ||
==Solution== | ==Solution== | ||
− | { | + | To factor <math>(n+1)(n+2)(n+3)(n+4) - 120</math>, we should try to find a way to create a quadratic in disguise. There, in fact, is a way! |
+ | |||
+ | Expand <math>(n+1)(n+4)</math> and <math>(n+2)(n+3)</math> separately: | ||
+ | |||
+ | <math>(n+1)(n+4)(n+2)(n+3) = (n^2+5n+4)(n^2+5n+6)</math> | ||
+ | |||
+ | We notice that there is a <math>n^2 + 5n</math> in both of these terms! Treat <math>n^2 + 5n</math> as a single quantity, <math>x</math>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. |
Revision as of 20:20, 30 June 2011
Problem
AoPSWiki:Problem of the Day/July 1, 2011
Solution
To factor , we should try to find a way to create a quadratic in disguise. There, in fact, is a way!
Expand and separately:
We notice that there is a in both of these terms! Treat as a single quantity, .
Our original quantity was .
We can factor the original polynomial as
Notice that can be factored as .
Our final factorization is .