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### Remainder and Factor Theorem

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Date: 01/29/98 at 08:09:46
From: Yanzhen
Subject: Remainder and Factor Theorem

Dear Dr. Maths,

How do I answer the following questions?

1) 2x^4 - 3x^3 - 18x^2 - 7 divided by (x^2) -1

2) When the polynomial p(x) is divided by (x-1), the remainder is 5,
and when p(x) is divided by (x-2), the remainder is 7.
Find the remainder, when p(x) is divided by (x-1)(x-2).

Thank you Dr Maths!
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Date: 02/24/98 at 14:30:13
From: Doctor Sonya
Subject: Re: Remainder and Factor Theorem

Dear Yanzhen,

To solve this problem, there is an important theorem you need to know,
called the Remainder and Factor theorem.  As I see from the subject
line, you've already heard of this, but I'll restate it anyway.

The theorem says that when you divide a polynomial f(x) by another one
g(x), there is a quotient q(x) and a remainder r(x).
These are related by the equation

f(x) = g(x)q(x) + r(x)

The remainder r(x) either is zero, or has degree lower than g(x).

For example, if g(x) is a linear polynomial, then r(x) should be a
constant, and if g(x) is quadratic, then r(x) should be a linear
polynomial of the form Ax+B.

I'm not going to prove this theorem, but if you want more information
on it, take a look in the Dr. Math archives, or write us back.

All right, now that this theorem's been stated, here's how we'll use

(1) Write  f(x) =  2x^4 - 3x^3 - 18x^2 - 7.  If the remainder is
Ax+B when this is divided by x^2 - 1 = (x-1)(x+1), we have

f(x) = (x-1)(x+1)q(x) + Ax + B.

This may not look like we've simplified anything, but wait,
what happens if you plug x = 1 into both sides of the equation?
You get:

f(1) = (1-1)(1+1)q(1) + 1A + B

Now evaluate both sides and see what you get.  Try the same
thing with x = -1.  Doing this will give you two equations for
A and B, which you can then solve.  Once you have A and B, plug
their values back into the remember Ax + B, and you'll be done.

(2) For your second problem, I'm going to write out what the
problem says in terms of the remainder and factor theorem.

p(x) = (x-1)q(x) + 5
p(x) = (x-2)Q(x) + 7

(Notice here that q(x) and Q(x) are different quotients.)

Now, try plugging 1 into the first equation and 2 into the
second equation.  You'll get that:

p(1) = 5 and that
p(2) = 7 .

Now, suppose the remainder is Ax+B when p(x) is divided by (x-1)(x-2).
We write

p(x) = (x-1)(x-2)q(x) + Ax + B.

Here, q(x) is the quotient. Putting x = 1 and x = 2 respectively
will give you two equations for A and B, which you can then solve
simultaneously to find out what Ax + B is.  When you plug in,
remember that you already know what p(1) and p(2) are.

-Doctors Sonya and Yiu,  The Math Forum
Check out our Web site  http://mathforum.org/dr.math/
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Associated Topics:
High School Polynomials

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