**Factorization** is the decomposition of an expression into a product of its factors.

The following are common factorizations.

For any positive integer $n$, $a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).$ In particular, for $n=2$, we have $a^2-b^2=(a-b)(a+b)$.

For $n$ an odd positive integer, $a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).$

$a^2 \pm 2ab + b^2 = (a\pm b)^2$

$x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$

$(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)$. $(ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)$.

$x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)$.

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

## 1. Find all ordered pairs of integer solutions $(x,y)$ such that $2^x+ 1 = y^2$.

Solution: We have $2^x = y^2-1 = (y-1)(y+1)$. Since the factors $(y-1)$ and $(y+1)$ on the right hand side are integers whose product is a power of 2, both $(y-1)$ and $(y+1)$ must be powers of 2. Furthermore, their difference is

$(y+1)-(y-1)=2,$

implying the factors must be $y+1 = 4$ and $y-1 = 2$. This gives $y=3$, and thus $x=3$. Therefore, $(3, 3)$ is the only solution.

## 2. Factorize the polynomial

$f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).$

Solution: Observe that if $a=b$, then $f(a, a, c) =0$; if $b=c$, then $f(a, b, b)=0$; and if $c=a$, then $f(c,b,c)=0$. By the Remainder-Factor Theorem, $(a-b), (b-c),$ and $(c-a)$ are factors of $f(a,b,c)$. This allows us to factorize

$f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).$

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