Difference between revisions of "Factoring"

Factoring is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an integer or polynomial) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.

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 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,

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

Vieta's/Newton Factorizations

These factorizations are useful for problems 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.

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

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 $(x-y)^3 + (y-z)^3 + (z-x)^3$.