REMAINDER THEOREM

The Remainder Theorem
The Remainder Theorem is useful for evaluating polynomials at a given
value of x, though it might not seem so, at least at first blush. This
is because the tool is presented as a theorem with a proof, and you
probably don't feel ready for proofs at this stage in your studies.
Fortunately, you don't "have" to understand the proof of the Theorem;
you just need to understand how to use the Theorem.

The Remainder Theorem starts with an unnamed polynomial p(x), where
"p(x)" just means "some polynomial p whose variable is x". Then the
Theorem talks about dividing that polynomial by some linear factor x –
a, where a is just some number. Then, as a result of the long
polynomial division, you end up with some polynomial answer q(x) (the
"q" standing for "the quotient polynomial") and some polynomial
remainder r(x).

As a concrete example of p, a, q, and r, let's look at the polynomial
p(x) = x3 – 7x – 6, and let's divide by the linear factor x – 4 (so a
= 4):

completed division: quotient x^2 + 4x + 9, remainder 30

So we get a quotient of q(x) = x2 + 4x + 9 on top, with a remainder of
r(x) = 30.

You know, from long division of regular numbers, that your remainder
(if there is one) has to be smaller than whatever you divided by. In
polynomial terms, since we're dividing by a linear factor (that is, a
factor in which the degree on x is just an understood "1"), then the
remainder must be a constant value. That is, when you divide by "x –
a", your remainder will just be some number.

The Remainder Theorem then points out the connection between division
and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3 = 12.
If you get a remainder, you do the multiplication and then add the
remainder back in. For instance, since 13 ÷ 5 = 2 R 3, then 13 = 5 × 2
+ 3. This process works the same way with polynomials. That is:

If p(x) / (x – a) = q(x) with remainder r(x),

then p(x) = (x – a) q(x) + r(x).

(Technically, this "if - then" statement is the "Division Algorithm
for Polynomials". But the Algorithm is the basis for the Remainder
Theorem.)

In terms of our concrete example: Copyright © Elizabeth Stapel
2002-2011 All Rights Reserved

Since (x^3 – 7x – 6) / (x – 4) = x2 + 4x + 9 with remainder 30,

then x3 – 7x – 6 = (x – 4) (x2 + 4x + 9) + 30.

The Remainder Theorem says that we can restate the polynomial in terms
of the divisor, and then evaluate the polynomial at x = a. But when x
= a, the factor "x – a" is just zero! Then evaluating the polynomial
at x = a gives us:

p(a) = (a – a)q(a) + r(a)
= (0)q(a) + r(a)
= 0 + r(a)
= r(a)

But remember that the remainder term r(a) is just a number! So the
value of the polynomial p(x) at
x = a is the same as the remainder you get when you divide that
polynomial p(x) by x – a. In terms of our concrete example:

p(4) = (4 – 4)((4)2 + 4(4) + 9) + 30
= (0)(16 + 16 + 9) + 30
= 0 + 30
= 30

But you gotta think: Okay, fine; the value of the polynomial p(x) at x
= a is the remainder r(a) when you divide by x – a, but who wants to
do the long division each time you have to evaluate a polynomial at a
given value of x?!? You're right; this would be overkill. Fortunately,
that's not what they really want you to do.

When you are dividing by a linear factor, you don't "have" to use long
polynomial division; instead, you can use synthetic division, which is
much quicker. In our example, we would get:

completed division: bottom row: 1 4 9 30

Note that the last entry in the bottom row is 30, the remainder from
the long division (as expected) and also the value of p(x) = x3 – 7x
– 6 at x = 4. And that is the point of the Remainder Theorem: There is
a simpler, quicker way to evaluate a polynomial p(x) at a given value
of x, and this simpler way is not to evaluate p(x) at all, but to
instead do the synthetic division at that same value of x. Here are
some examples:

* Use the Remainder Theorem to evaluate f (x) = 6x3 – 5x2 + 4x –
17 at x = 3.

First off, even though the Remainder Theorem refers to the
polynomial and to long division and to restating the polynomial in
terms of a quotient, a divisor, and a remainder, that's not actually
what I'm meant to be doing. Instead, I'm supposed to be doing
synthetic division, using "3" as the divisor:

completed division: bottom row: 6 13 43 112

Since the remainder (the last entry in the bottom row) is 112,
then the Remainder Theorem says that:

f (3) = 112.

* Using the Remainder Theorem, find the value of f (–5), for f
(x) = 3x4 + 2x3 + 4x.

I need to do the synthetic division, remembering to put zeroes
in for the powers of x that are not included in the polynomial:

completed division: bottom row: 3 -13 65 -321 1605

Since the remainder is 1605, then, thanks to the Remainder
Theorem, I know that:

f (–5) = 1605.

* Use the Remainder Theorem to determine whether x = 2 is a zero of
f (x) = 3x7 – x4 + 2x3 – 5x2 – 4

For x = 2 to be a zero of f (x), then f (2) must evaluate to
zero. In the context of the Remainder Theorem, this means that my
remainder, when dividing by x = 2, must be zero:

completed division: bottom row: 3 6 12 23 48 91 182 360

The remainder is not zero. Then x = 2 is not a zero of f (x).

* Use the Remainder Theorem to determine whether x = –4 is a solution of
x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8 = 0

For x = –4 to be a solution of f (x) = x6 + 5x5 + 5x4 + 5x3 +
2x2 – 10x – 8 = 0, it must be that f (–4) = 0. In the context of the
Remainder Theorem, this means that the remainder, when dividing by x =
–4, must be zero:

completed division: bottom row: 1 1 1 1 -2 -2 0

The remainder is zero. Then x = –4 is a solution of the given equation.