Polynomial remainder theorem

Not to be confused with Bézout's theorem.

In algebra, the polynomial remainder theorem or little Bézout's theorem^[1] is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-a$ is equal to $f(a).$ In particular, $x-a$ is a divisor of $f(x)$ if and only if $f(a)=0.$ ^[2]

Examples

Example 1

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

Example 2

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

{\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

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

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

Proof

The polynomial remainder theorem follows from the definition 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

f(x)=q(x)g(x)+r(x)\quad {\text{and}}\quad r(x)=0\;{\text{or}}\;\deg(r)<\deg(g)\,.

If we take $g(x)=x-a$ as the divisor, either r = 0 or its degree is zero; in both cases, r is a constant that is independent of x; that is

f(x)=q(x)(x-a)+r\,.

Setting $x=a$ in this formula, we obtain:

f(a)=r\,.

A slightly different and slightly more elementary proof starts with an observation that $f(x)-f(a)$ is a linear combination of the terms of the form $x^{k}-a^{k}$ , each of which is divisible by $x-a$ since $x^{k}-a^{k}=(x-a)(x^{k-1}+x^{k-2}a+\dots +xa^{k-2}+a^{k-1})$ .

Applications

The polynomial remainder theorem may be used to evaluate $f(a)\,$ by calculating the remainder, $r$ . 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

↑ Piotr Rudnicki (2004). "Little Bézout Theorem (Factor Theorem)" (PDF). Formalized Mathematics. 12 (1): 49–58.
↑ Larson, Ron (2014), College Algebra, Cengage Learning
↑ Larson, Ron (2011), Precalculus with Limits, Cengage Learning

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