Polynomial remainder theorem

In algebra, the polynomial remainder theorem or little Bézout's theorem (named after Étienne Bézout)<ref>Template:Cite journal</ref> is an application of Euclidean division of polynomials. It states that, for every number <math>r</math>, any polynomial <math>f(x)</math> is the sum of <math>f(r)</math> and the product of <math>x-r</math> and a polynomial in <math>x</math> of degree one less than the degree of <math>f</math>. In particular, <math>f(r)</math> is the remainder of the Euclidean division of <math>f(x)</math> by <math>x-r</math>, and <math>x-r</math> is a divisor of <math>f(x)</math> if and only if <math>f(r)=0</math>,<ref>Larson, Ron (2014), College Algebra, Cengage Learning</ref> a property known as the factor theorem.

Examples

Example 1

Let <math>f(x) = x^3 - 12x^2 - 42</math>. Polynomial division of <math>f(x)</math> by <math>(x-3)</math> gives the quotient <math>x^2 - 9x - 27</math> and the remainder <math>-123</math>. By the polynomial remainder theorem, <math>f(3)=-123</math>.

Example 2

Proof that the polynomial remainder theorem holds for an arbitrary second degree polynomial <math>f(x) = ax^2 + bx + c</math> by using algebraic manipulation: <math display="block">\begin{align} f(x)-f(r)

&= ax^2+bx+c-(ar^2+br+c)\\
&= a(x^2-r^2)+ b(x-r)\\
&= a(x-r)(x+r)+b(x-r)\\
&= (x-r)(ax +ar+ b)

\end{align}</math>

So, <math display="block">f(x) = (x - r)(ax + ar + b) + f(r), </math> which is exactly the formula of Euclidean division.

The generalization of this proof to any degree is given below in Template:Slink.

Proofs

Using Euclidean division

The polynomial remainder theorem follows from the theorem of Euclidean division, which, given two polynomials Template:Math (the dividend) and Template:Math (the divisor), asserts the existence (and the uniqueness) of a quotient Template:Math and a remainder Template:Math such that

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

If the divisor is <math>g(x) = x-r,</math> where r is a constant, then either Template:Math or its degree is zero; in both cases, Template:Math is a constant that is independent of Template:Math; that is

<math>f(x)=Q(x)(x-r) + R.</math>

Setting <math>x=r</math> in this formula, we obtain:

<math>f(r)=R.</math>

Direct proof

A constructive proofTemplate:Mdashthat does not involve the existence theorem of Euclidean divisionTemplate:Mdashuses the identity

<math>x^k-r^k=(x-r)(x^{k-1}+x^{k-2}r+\dots+xr^{k-2}+r^{k-1}).</math>

If <math>S_{k}</math> denotes the large factor in the right-hand side of this identity, and

<math>f(x)=a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x +a_0,</math>

one has

<math>f(x)-f(r)=(x-r)(a_n S_n +\cdots + a_2S_2 +a_1),</math>

(since <math>S_1=1</math>).

Adding <math>f(r)</math> 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 <math>f(r)</math> by calculating the remainder, <math>R</math>. 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.<ref>Larson, Ron (2011), Precalculus with Limits, Cengage Learning</ref>