Difference between revisions of "Factoring"
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<math>a^n-b^n=(a-b)(a^{n-1}+ba^{n-2} + \cdots + b^{n-2}a + b^{n-1})</math> | <math>a^n-b^n=(a-b)(a^{n-1}+ba^{n-2} + \cdots + b^{n-2}a + b^{n-1})</math> | ||
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<math>a^n+b^n=(a+b)(a^{n-1} - ba^{n-2} + b^2a^{n-3} - b^3a^{n-4} + \cdots + b^{n-1})</math> | <math>a^n+b^n=(a+b)(a^{n-1} - ba^{n-2} + b^2a^{n-3} - b^3a^{n-4} + \cdots + b^{n-1})</math> |
Revision as of 03:25, 23 June 2006
Note to readers and editers: Please fix up this page by adding in material from Joe's awesome factoring page.
Contents
Why Factor
Factoring equations is an essential part of problem solving. Applying number theory to products yields many results.
There are many ways to factor.
Differences and Sums of Powers
Using the formula for the sum of a geometric sequence, it's easy to derive the more general formula:
Take note of the specific case where n is odd:
This also leads to the formula for the sum of cubes,
Simon's Trick
See Simon's Favorite Factoring Trick (This is not a recognized formula, please do not quote it on the USAMO or similar national proof contests)
Summing Sequences
Also, it is helpful to know how to sum arithmetic sequence and geometric sequence.
Vieta's/Newton Factorizations
These factorizations are useful for problem that could otherwise be solved by Newton sums or problems that give a polynomial, and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere.
Another Useful Factorization
Practice Problems
- Prove that is never divisible by 121 for any positive integer
- Prove that is divisible by 7 - USSR Problem Book
- Factor