Polynomial remainder theorem

In algebra, the polynomial remainder theorem or little Bézout's theorem (named after Étienne Bézout)^[1] is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial ${\displaystyle f(x)}$ by a linear polynomial ${\displaystyle x-r}$ is equal to ${\displaystyle f(r).}$ In particular, ${\displaystyle x-r}$ is a divisor of ${\displaystyle f(x)}$ if and only if ${\displaystyle f(r)=0,}$^[2] a property known as the factor theorem.

Examples[]

Example 1[]

Let ${\displaystyle f(x)=x^{3}-12x^{2}-42}$. Polynomial division of ${\displaystyle f(x)}$ by ${\displaystyle (x-3)}$ gives the quotient ${\displaystyle x^{2}-9x-27}$ and the remainder ${\displaystyle -123}$. Therefore, ${\displaystyle f(3)=-123}$.

Example 2[]

Show that the polynomial remainder theorem holds for an arbitrary second degree polynomial ${\displaystyle f(x)=ax^{2}+bx+c}$ by using algebraic manipulation:

${\displaystyle {\begin{aligned}{\frac {f(x)}{x-r}}&={\frac {a{x^{2}}+bx+c}{x-r}}\\&={\frac {a{x^{2}}-arx+arx+bx+c}{x-r}}\\&={\frac {ax(x-r)+(b+ar)x+c}{x-r}}\\&=ax+{\frac {(b+ar)(x-r)+c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {a{r^{2}}+br+c}{x-r}}\end{aligned}}}$

Multiplying both sides by (x − r) gives

${\displaystyle f(x)=ax^{2}+bx+c=(ax+b+ar)(x-r)+{a{r^{2}}+br+c}}$.

Since ${\displaystyle R=ar^{2}+br+c}$ is the remainder, we have indeed shown that ${\displaystyle f(r)=R}$.

Proof[]

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

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

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:

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

A slightly different proof, which may appear to some people as more elementary, starts with an observation that ${\displaystyle f(x)-f(r)}$ is a linear combination of terms of the form ${\displaystyle x^{k}-r^{k},}$ each of which is divisible by ${\displaystyle x-r}$ since ${\displaystyle x^{k}-r^{k}=(x-r)(x^{k-1}+x^{k-2}r+\dots +xr^{k-2}+r^{k-1}).}$

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.^[3]

References[]