# The Remainder Theorem

• Sep 23rd 2011, 02:06 PM
Dragon08
The Remainder Theorem
The remainder theorem says when a polynomial function is divided by ax-b, the remainder is f(b/a). But if you take a function and divided it by 2x-1, for example, the remainder will be obviously different than when divided by 4x-2, which is 2x-1 multiplied by two. f(2/4) is the same as f(1/2), so the result when subbed into f(x) should be the same, but their remainders are obviously different. Can anybody clarify?
• Sep 23rd 2011, 10:16 PM
CaptainBlack
Re: The Remainder Theorem
Quote:

Originally Posted by Dragon08
The remainder theorem says when a polynomial function is divided by ax-b, the remainder is f(b/a). But if you take a function and divided it by 2x-1, for example, the remainder will be obviously different than when divided by 4x-2, which is 2x-1 multiplied by two. f(2/4) is the same as f(1/2), so the result when subbed into f(x) should be the same, but their remainders are obviously different. Can anybody clarify?

P(x)=(4x-2)Q(x)+R

and

P(x)=(2x-1)Q'(x)+R'

Q'(x)=2Q(x)

and R=R'

CB
• Sep 24th 2011, 06:29 AM
Dragon08
Re: The Remainder Theorem
How did you know that Q'(x)=2Q(x)?
• Sep 24th 2011, 06:38 AM
CaptainBlack
Re: The Remainder Theorem
Quote:

Originally Posted by Dragon08
How did you know that Q'(x)=2Q(x)?

When doing polynomial long division of P(x) by (4x-2) and by (2x-1) the quotients differ by a factor of 2.

Also:

P(x)=(4x-2)Q(x)+R=(2x-1) [2Q(x)] +R

But R is a pure number, so:

Q'(x)=2Q(x) and R=R'

CB
• Sep 24th 2011, 06:46 AM
Dragon08
Re: The Remainder Theorem
Oh, okay. Thanks. So this only works for fraction divisors right? For example, 10 divided by 3 and 10 divided by 6 don't get the same remainders.