Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 1, 2011"
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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>. | ||
− | + | <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>. | Our final factorization is <math>\boxed{(n-1)(n+6)(n^2+5n+16)}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 10:57, 1 July 2011
Problem
AoPSWiki:Problem of the Day/July 1, 2011
Solution 1
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 .
can be factored as .
Our final factorization is .
Solution 2
Let . Then and (since ). Therefore I have: where is a quadratic. Simplifying yields as before, which is irreducible as .