
The polynomial remainder theorem states, If f(x) is a polynomial, then the remainder obtained by dividing f(x) by xr equals f(r) In other words, xr is a factor of f(x)f(r) To show this is true, we will first show that xr is a factor of ax^{n}ar^{n}, 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 ax^{n}ar^{n}, and xr divides every term of that sum, then it follows that xr 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 (xr)  (ax^{n}ar^{n}) for all real a and nonnegative integer n. If (xr)  (x^{n}r^{n}) then (xr)  (ax^{n}ar^{n}), so I'll just show the former. If n=0, then xr is a factor of x^{n}r^{n}, because x^{0}r^{0}=0, and xr is a factor of 0. Now let's suppose xr is a factor of x^{n}r^{n}. We will show xr is a factor of x^{n+1}r^{n+1}. First, it is clear that (xr)  (x^{n}r^{n})(xr). I will write a series of expressions that are all equivalent to (x^{n}r^{n})(xr).
This proves that (xr)  (x^{n}r^{n}) for all nonnegative integer n. Since the polynomial f(x)f(r) is composed of a sum of numbers of the form ax^{n}ar^{n}, and xr divides every term of that sum, then it follows that xr divides f(x)f(r). In other words, If f(x) is a polynomial, then the remainder obtained by dividing f(x) by xr equals f(r) 
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