Answer:

3.2 The Factor Theorem and The Remainder Theorem 257

3.2 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 polynomials.

1

and probably forgot

2pun inte

258 Polynomial Functions

Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where

the degree of p is greater than or equal to the degree of d. There exist two unique polynomials,

q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is

strictly less than the degree of d.

As you may recall, all of the polynomials in Theorem 3.4 have special names. The polynomial p

is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then

d is called a factor of p. The proof of Theorem 3.4 is usually relegated to a course in Abstract

Algebra,3 but we can still use the result to establish two important facts which are the basis of the

rest of the chapter.

Theorem 3.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1

and c is a real number. When p(x) is divided by x − c the remainder is p(c).

The proof of Theorem 3.5 is a direct CONSEQUENCE of Theorem 3.4. When a polynomial is divided

by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree

1, the degree of the remainder must be 0, which means the remainder is a constant. Hence, in

either case, p(x) = (x − c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows

that p(c) = (c − c) q(c) + r = 0 · q(c) + r = r, so we get r = p(c) as required. There is one last ‘low

hanging fruit’4

to collect which we present below.

Theorem 3.6. The Factor Theorem: Suppose p is a nonzero polynomial. The real number

c is a zero of p if and only if (x − c) is a factor of p(x).

The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a factor

of p(x), this means p(x) = (x − c) q(x) for some polynomial q. Hence, p(c) = (c − c) q(c) = 0, so c

is a zero of p. CONVERSELY, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem

tells us the remainder when p(x) is divided by (x − c), namely p(c), is 0, which means (x − c) is a

factor of p. What we have established is the fundamental connection between zeros of polynomials

and factors of polynomials.

Of the things The Factor Theorem tells us, the most PRAGMATIC is that we had better find a more

efficient way to divide polynomials by quantities of the form x − c. Fortunately, people like Ruffini

and Horner have already blazed this trail. Let’s take a closer look at the long division we performed

at the beginning of the section and try to streamline it. First off, let’s change all of the subtractions

into ADDITIONS by distributing through the −1s.Step-by-step explanation: