Applying the Remainder Theorem

Date: 10/25/1999 at 08:37:07
From: Aimee
Subject: Remainder Theorem

I have tried this question three times, different ways, and for some 
reason I always get mixed up (it could be with my f(x) etc.)

The expression 6x^2 + x + 7 leaves the same remainder when divided 
by (x-a) and by (x+2a) where a is not equal to 0. Calculate the value 
of a.

We have been through it in class but I am studying for a test and 
don't understand what I have been doing. If you can explain it, it 
would be a great help!

Thanks,
Aimee

Date: 10/25/1999 at 18:03:02
From: Doctor Schwa
Subject: Re: Remainder Theorem

Aimee,

This is a pretty clever question!

The remainder theorem tells us that:

The remainder when you divide a polynomial by (x-a) is the same as 
what you get if you plug x = a into the polynomial.

For instance, if I divide x^8 + 3x^2 - 7x + 1 by (x-2), I get a 
remainder of 2^8 + 3*2^2 - 7*2 + 1 = 255.

In your problem, when you divide by (x-a) the remainder will be

     6a^2 + a + 7

and when you divide by (x + 2a), that's the same as plugging in -2a,

     6 (-2a)^2 + (-2a) + 7

Setting those two things equal to each other should give you an 
equation to solve for a. One of the solutions will be a = 0, but then 
there will be another more interesting solution.

I hope that helps clear things up.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/