# Difference between revisions of "Factoring"

### 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See Simon's Favorite Factoring Trick (This is not a recognized formula, please do not quote it on the USAMO or similar national proof contests)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 $(x-y)^3 + (y-z)^3 + (z-x)^3$