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: 
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
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 root (zero) of f (x)
Let's take a look at some example questions:
Find the remainder when 4x^{2} + 2x  5 is divided by (x  1). 
Using the Remainder Theorem, we can find this answer quickly.
Think of f ( x) = 4 x^{2} + 2 x  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. 
Is (x + 4) a factor of x^{4} + 6x^{3} + 7x^{2}  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.


Given P(x) = x^{3} + 3x^{2} + 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) = 3^{3} + 3(3)^{2} + a(3)  15
0 = 3^{3} + 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.
