When working with Synthetic Division, we saw a series of division problems involving a divisor of the
form x - a , where the degree of the divisor was one. When you divide by a polynomial of degree one
(such as x - a), the remainder will have a degree of 0. The remainder will be a constant, such as "2").

Division Algorithm for Polynomials

If we multiply each side of the Division Algorithm by the divisor, d(x), we have:

Now, if we use this new equation when working with a divisor of x - a, we have:
f (x) = (x - a)•q(x) + r(x)
but, r(x) is simply a constant, since we are dividing by a degree of 1, so replace r(x) with just r:
f (x) = (x - a)•q(x) + r
Now, when x = a, we have:
f (a) = (a - a)•q(a) + r
f (a) = (0)•q(a) + r
f (a) = r
We now have the basis of the Remainder Theorem for Polynomials:

The Polynomial Remainder Theorem: When the polynomial f (x) is divided by x - a, the remainder equals f (a).

Great discovery!! Now, when you divide a polynomial, f (x), by x - a, you won't need to actually do the division to find the remainder. Simply calculate f (a). Plug a into f (x) and the answer is the remainder.

The most valuable use of this discovery is to determine if the divisor (x - a) is a factor of the dividend. If (x - a) is a factor of f (x), the remainder will be zero. You can quickly make this determination by plugging a into f (x) to see if the result is zero.

This special use of the Remainder Theorem to determine a factor is call the Factor Theorem:

The Factor Theorem: If f (a) = 0 for polynomial f (x), then x - a must be a factor of the polynomial. Converse: If x - a is a factor of polynomial f (x), then f (a) = 0.

The factor theorem links factors and roots (zeros) of a polynomial.
factor (x - 2);        f (2) = 0;        2 is a zero (root) of f (x)

Let's take a look at some example questions:

Find the remainder when 4x2 + 2x - 5 is divided by (x - 1).

Using the Remainder Theorem, we can find this answer quickly. Think of f (x) = 4x2 + 2x - 5 and solve for f (1). Notice that we are substituting "1", the root value associated with (x - 1). f (1) = 4(1)2 + 2(1) - 5 = 1                The remainder is 1.