Factorization of polynomials

A polynomial factorization expresses a polynomial as a product of other polynomials. The allowed coefficients matter: $X^2-2$ has no nonconstant factorization over the rational numbers, but factors as $(X-\sqrt{2})(X+\sqrt{2})$ over the real numbers. Difference of squares gives $X^2-Y^2=(X-Y)(X+Y)$ over any commutative ring.

For $n\ge 1$ and $a,b$ in a commutative ring, multiplying out the displayed polynomial factorization cancels all intermediate terms, leaving $a^n-b^n$. The case $n=2$ is the difference of squares. Unlike division by $a-b$, this identity also applies when $a=b$.

For real $1\le b\le a$, every summand lies between one and $a^{n-1}$. Consequently $n(a-b)\le a^n-b^n\le n(a-b)a^{n-1}$. This turns a one-step discrepancy into a bound after repeated multiplication, for example when estimating the global error of a one-step method on $y^{\prime }=y$.

A nonzero polynomial over a field is square-free if no square of a nonconstant irreducible polynomial divides it. Equivalently, in its factorization of polynomials, every irreducible polynomial factor occurs with exponent one. For example, $x(x^2+1)$ is square-free over the real numbers, whereas $(x-1{)}^2$ is not. Square-freeness over a given field concerns factors in that field; over the real numbers or complex numbers, it is equivalent to having no repeated polynomial roots after splitting over the complex numbers.

A sieve that marks reducible polynomials by their divisibility by smaller irreducible polynomials. It is the polynomial-ring analogue of a number-theoretic sieve.

Replacing $b$ by $-b$ in the difference of two cubes givesFor real cube roots $u=a^{1/3}$ and $v=b^{1/3}$, this simplifies $(a^{1/3}+b^{1/3})(a^{2/3}-(ab{)}^{1/3}+b^{2/3})$ to $a+b$. Specifying real cube roots matters when using such identities with negative real inputs; unconstrained complex exponentiation has logarithm-branch choices.

The polynomial identityfollows by multiplying out the right side. For real $a,b$, the quadratic factor is $a^2+ab+b^2=(a+b/2{)}^2+3b^2/4$, so it is nonnegative and vanishes only when $a=b=0$. The identity is a special case of the finite geometric series factorisation $a^n-b^n=(a-b){\sum }_{j=0}^{n-1}{a}^{n-1-j}{b}^j$.