Polynomial remainder theorem

Proofs

Using Euclidean division

The polynomial remainder theorem follows from the theorem of Euclidean division, which, given two polynomials f(x) (the dividend) and g(x) (the divisor), asserts the existence (and the uniqueness) of a quotient Q(x) and a remainder R(x) such that

${\displaystyle f(x)=Q(x)g(x)+R(x)\text{and}R(x)=0\text{ or }\mathrm{deg}(R)<\mathrm{deg}(g).}$

If the divisor is ${\displaystyle g(x)=x-r,}$ where r is a constant, then either R(x) = 0 or its degree is zero; in both cases, R(x) is a constant that is independent of x; that is

${\displaystyle f(x)=Q(x)(x-r)+R.}$

Setting ${\displaystyle x=r}$ in this formula, we obtain:

${\displaystyle f(r)=R.}$

Direct proof

A constructive proof—that does not involve the existence theorem of Euclidean division—uses the identity

${\displaystyle x^k-r^k=(x-r)(x^{k-1}+x^{k-2}r+\cdots +x{r}^{k-2}+r^{k-1}).}$

If ${\displaystyle S_k}$ denotes the large factor in the right-hand side of this identity, and

${\displaystyle f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0,}$

one has

${\displaystyle f(x)-f(r)=(x-r)(a_n{S}_n+\cdots +a_2{S}_2+a_1),}$

(since ${\displaystyle S_1=1}$).

Adding ${\displaystyle f(r)}$ to both sides of this equation, one gets simultaneously the polynomial remainder theorem and the existence part of the theorem of Euclidean division for this specific case.

Applications

The polynomial remainder theorem may be used to evaluate ${\displaystyle f(r)}$ by calculating the remainder, ${\displaystyle R}$. Although polynomial long division is more difficult than evaluating the function itself, synthetic division is computationally easier. Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem.

The factor theorem is another application of the remainder theorem: if the remainder is zero, then the linear divisor is a factor. Repeated application of the factor theorem may be used to factorize the polynomial.