$p(x)$ has degree 4 and $g(x)$ has degree 2. If $p(x)=q(x)\cdot g(x)+r(x)$ with $r(x)$ ≠ 0, which pair of degrees is possible?

**Theorem**: For polynomials $p(x)$ (dividend) and $g(x)$ ≠ 0 (divisor), there exist UNIQUE polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that

$p(x)=q(x)\cdot g(x)+r(x)$,

where either $r(x)=0$ or deg $r(x)$ < deg $g(x).\ast \ast$Mechanism**: long division of polynomials, exactly analogous to integer long division.

**Use case**: if you know one zero (or factor) of a higher-degree polynomial, divide it out to get a lower-degree quotient — then factor that.

**Example**: $p(x)=x^3-3x^2-x+3$. Try $x=1:p(1)=0$ → $(x-1)$ is a factor. Divide: $p(x)÷(x-1)=x^2-2x-3=(x-3)(x+1)$. So $p(x)=(x-1)(x-3)(x+1)$.