The Factor Theorem and The Remainder Theorem

Suppose we wish to FIND the zeros of f(x) = x

3 + 4x

2 − 5x − 14. Setting f(x) = 0 results in the

polynomial equation x

3 + 4x

2 − 5x − 14 = 0. Despite all of the FACTORING techniques we learned1

in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing

calculator, we GET

The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show

that f(2) = 0, but the other two zeros seem to be less friendly. Even though we could use the

‘Zero’ command to find decimal approximations for these, we seek a method to find the remaining

zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor

of (x − 2) lurking AROUND in the factorization of f(x). In other words, we should expect that

x

3 + 4x

2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we find such

a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing

x

3 + 4x

2 − 5x − 14 by x − 2 gives

x

2 + 6x + 7

x−2 x

3 + 4x

2 − 5x − 14

−

x

3 −2x

2

6x

2 − 5x

−

6x

2 −12x)

7x − 14

− (7x −14)

0

As you may recall, this means x

3 + 4x

2 − 5x − 14 = (x − 2)

x

2 + 6x + 7

, so to find the zeros of f,

we now solve (x − 2)

x

2 + 6x + 7

= 0. We get x − 2 = 0 (which gives us our known zero, x = 2)

as well as x

2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic Formula to

get x = −3 ±

√

2. The point of this section is to generalize the technique applied here. First up is

a friendly reminder of what we can expect when we divide polynomia