Factoring

Revision as of 04:20, 23 June 2006 by Calculuslover800 (talk | contribs) (Differences and Sums of Powers: got dots to show up right)

Note to readers and editers: Please fix up this page by adding in material from Joe's awesome factoring page.


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

$a^2-b^2=(a+b)(a-b)$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Using the formula for the sum of a geometric sequence, it's easy to derive the more general formula:

$a^n-b^n=(a-b)(a^{n-1}+ba^{n-2} + \cdots + b^{n-2}a + b^{n-1})$

Take note of the specific case where $b$ is negative and n is odd:

$a^n+b^n=(a+b)(a^{n-1} - ba^{n-2} + b^2a^{n-3} - b^3a^{n-4} + \cdots + b^{n-1})$

This also leads to the formula for the sum of cubes,

$a^3+b^3=(a+b)(a^2-ab+b^2)$

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See Simon's Favorite Factoring Trick (This is not a recognized formula, please do not quote it on the USAMO or similar national proof contests)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.

  • $\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
  • $\displaystyle (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$
  • $\displaystyle (a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$

Another Useful Factorization

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Practice Problems

  • Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$
  • Prove that $2222^{5555} + 5555^{2222}$ is divisible by 7 - USSR Problem Book
  • Factor $(x-y)^3 + (y-z)^3 + (z-x)^3$

Other Resources

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