# 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

Using the formula for the sum of a geometric sequence, it's easy to derive the more general formula: $\[a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})\]$

If $n=2$, this creates the difference of squares factorization, $\[a^2-b^2=(a+b)(a-b)\]$

This leads to the difference of cubes factorization, $\[a^3-b^3=(a-b)(a^2+ab+b^2)\]$

In addition, if $n$ is odd: $\[a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})\]$

This also leads to the formula for the sum of cubes, $\[a^3+b^3=(a+b)(a^2-ab+b^2)\]$

Another way to discover these factorizations is the following: the expression $a^n - b^n$ is equal to zero if $a = b$. If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so $a^n - b^n$ must have a factor of $a - b$. Similarly, we note that the expression $a^n + b^n$ when $n$ is odd is equal to zero if $a = -b$, so it must have a factor of $a - (-b) = a + b$. Note that when $n$ is even, $(-b)^n + b^n = 2b^n$, rather than 0, so this gives us no useful information.

## 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.

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

## 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 $(x-y)^3 + (y-z)^3 + (z-x)^3$.
• Factor $x^4 + 1$ into two polynomials with real coefficients.
• Given that $a+b+c=0$, prove that $abc=\dfrac{a^3+b^3+c^3}{3}$.

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