factor theorem

Factor theorem

For a polynomial $p$ and a number $r$:

$$(x-r)\mid p(x)⟺p(r)=0$$

That is, $p(x)=(x-r)q(x)$ for some polynomial $q$ of degree $\mathrm{deg}p-1$, exactly when $r$ is a root of $p$.
Roots and linear factors are the same thing; factoring a polynomial means finding its roots.

Polynomial division with remainder

Polynomials divide like integers: for a polynomial $p$ and divisor $d\ne 0$ there are unique polynomials $q$ (quotient) and $\rho$ (remainder) with

$$p(x)=q(x)d(x)+\rho (x),\mathrm{deg}\rho <\mathrm{deg}d$$

Same shape as $17=5\cdot 3+2$ with remainder $2<3$.

Proof:

Divide $p$ by $(x-r)$. The divisor has degree $1$, so the remainder has degree $0$, a constant $c$:

$$p(x)=q(x)(x-r)+c$$

Evaluate both sides at $x=r$. The first term vanishes:

$$p(r)=q(r)\underset{=0}{\underbrace{(r-r)}}+c=c$$

So the remainder of dividing by $(x-r)$ is the value $p(r)$.
”$(x-r)$ divides $p$” means remainder zero, which now reads $p(r)=0$. $■$

EXAMPLE

$p(x)=x^2-x-2$.
$p(2)=4-2-2=0$, so $(x-2)$ divides: indeed $x^2-x-2=(x-2)(x+1)$, remainder $0$.
$p(1)=1-1-2=-2\ne 0$, so $(x-1)$ does not divide: $x^2-x-2=(x-1)\cdot x-2$, remainder $-2=p(1)$.

Each root $r$ found splits off one factor, $p=(x-r)q$, and the search continues on $q$, whose degree is one lower.

A polynomial of degree $n$ therefore has at most $n$ roots: each costs one degree.
Over $\mathbb{C}$, we get exactly $n$ roots counting multiplicity, and the full factorization into linear factors: $a_n{\prod }_k(x-z_k)$ (fundamental theorem of algebra).
Over $\mathbb{R}$ it can stop early: quadratics without real roots, e.g. $x^2+1$, cannot be factored further, so we get a factorization into linear and irreducible quadratic factors.

The constant $a_n$ in front is the leading coefficient: every factor $(x-r)$ is monic (coefficient of $x$ is $1$), so multiplying out ${\prod }_k(x-z_k)$ gives $x^n+\dots$, and matching the $x^n$ coefficient of $p$ requires the factor $a_n$.

The lower coefficients need no constants of their own; the roots determine them.
Degrees of freedom match up: $n+1$ coefficients $=$ $n$ roots $+1$ scale $a_n$.

EXAMPLE

$$3x^2-3x-6=3(x^2-x-2)=3(x-2)(x+1)$$

For a polynomial $p$ and a number $r$, how does $(x-r)$ dividing $p$ relate to $p(r)$? $(x-r)$ divides $p$ if and only if $p(r)=0$ Roots correspond to linear factors; factoring means finding roots.