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 Skip Navigation LinksMath Help > Basic Algebra > Polynomial > Polynomial Remainder Theorem

 

The polynomial remainder theorem states,

If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-r equals f(r)

In other words, x-r is a factor of f(x)-f(r)

To show this is true, we will first show that x-r is a factor of axn-arn, where a is any real number, and n is a nonnegative integer.  Then, since the polynomial f(x)-f(r) is composed of a sum of numbers of the form axn-arn, and x-r divides every term of that sum, then it follows that x-r divides f(x)-f(r).


Before we begin, perhaps a definition of "remainder" is needed.  As you recall, in division of integers, if r is the remainder of the division p/d, then there is an integer, q, such that qd+r=p, and 0 <= r < d.  In division of polynomials the same idea holds, except the quotient and remainders are both polynomials.  In division of polynomials, if r is the remainder of the division p/d (both polynomials), then there is a polynomial, q, such that qd+r=p, and 0 <= O(r) < d.  Here O(r) means the "order" or "degree" of polynomial r.


Proof that (x-r) | (axn-arn) for all real a and nonnegative integer n.

If (x-r) | (xn-rn) then (x-r) | (axn-arn), so I'll just show the former.

If n=0, then x-r is a factor of xn-rn, because x0-r0=0, and x-r is a factor of 0.

Now let's suppose x-r is a factor of xn-rn.  We will show x-r is a factor of xn+1-rn+1.

First, it is clear that (x-r) | (xn-rn)(x-r).  I will write a series of expressions that are all equivalent to (xn-rn)(x-r).

(x-r) | (xn-rn)(x-r)
(x-r) | xn+1-xnr-rnx+rn+1     by multiplying the terms together
(x-r) | xn+1-xnr+2rn+1-rnx-rn+1    by adding and subtracting 2arn+1.
(x-r) | xn+1-xnr+rnr+rnr-rnx-rn+1    by expressing 2arn+1 as arnr+arnr.
(x-r) | xn+1-r(xn-rn)-(x-r)(rn)-rn+1     by gathering terms, and factoring r and (x-r)
(x-r) | (xn+1-rn+1) - r(xn-rn) - (x-r)(rn)   by rearranging terms

(x-r) | r(xn-rn)
(x-r) | (x-r)(rn)
(x-r) | (xn+1-rn+1)   (x-r)|a-b-c, (x-r)|b, (x-r)|c, so (x-r)|a.

This proves that (x-r) | (xn-rn) for all nonnegative integer n.
Therefore (x-r) | (axn-arn) for all real a and nonnegative integer n.


Since the polynomial f(x)-f(r) is composed of a sum of numbers of the form axn-arn, and x-r divides every term of that sum, then it follows that x-r divides f(x)-f(r).

In other words,

If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-r equals f(r)

Related pages in this website

Number Theory

Synthetic Division

Descartes' Rule Of Signs

Polynomial

 

The webmaster and author of this Math Help site is Graeme McRae.