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").

Before we continue, consider:
Division Algorithm for Polynomials
If we multiply each side of the Division Algorithm
by the divisor, d(x), we have:
division algornew
f (x) is the dividend; d(x) is the divisor; q(x) is the quotient; r(x) is the remainder

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
(a) = r
We now have the basis of the Remainder Theorem for Polynomials:

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

hint 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:

theorem The Factor Theorem: If f (a) = 0 for polynomial f (x), then x - a must be a factor of the polynomial.
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.
(x - 2);        f (2) = 0;        2 is a root (zero) of f (x)


Let's take a look at some example questions:

expin1 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.


expin2 Is (x + 4) a factor of   x4 + 6x3 + 7x2 - 6x - 8?
Remember that when a polynomial is divided by a "factor", the remainder is zero. We simply need to use the Remainder Theorem to see if the remainder is zero.
f (-4) = (-4)4 + 6(-4)3 +7(-4)2 - 6(-4) - 8
= 256+ (-384) + 112 - (-24) - 8 = 0
    The remainder is 0, so (x + 4) is a factor.


expin3 Given P(x) = x3 + 3x2 + ax - 15.
Find a if (x - 3) is a factor of P(x).
If (x - 3) is a factor of P(x), then P(3) will be zero. Set P(3) = 0 and solve for a.
P(3) = 33 + 3(3)2 + a(3) - 15
       0 = 33 + 3(3)2 + a(3) - 15
0 = 27 + 27 + 3a - 15
0 = 39 + 3a
-39 = 3a
-13 = a
       If (x - 3) is a factor of P(x), then a = -13.


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