# A tricky remainder theorem problem

## Homework Statement

A polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)

## Homework Equations

Remainder theorem

## The Attempt at a Solution

I know that

P(x) = (x-1)A(x) + 1
and P(x) = (x+1)B(x) + 3

But how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.

From your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ($x^2 - 1$) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation $$P(x) = (x^2 - 1)f(x) + Kx + L$$ . Try to proceed from here.

• SammyS
Why is the remainder is a linear expression? I can't catch your explanation.

Ray Vickson
Homework Helper
Dearly Missed
Why is the remainder is a linear expression? I can't catch your explanation.

The remainder in ##a(x)/b(x)## is a polynomial ##r(x)## of degree strictly less than the degree of ##b(x)##. Basically, that is the _definition_ of "remainder".

explanation

So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.

Ray Vickson
Homework Helper
Dearly Missed
So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.

So it has something to do with the divisor right?

rcgldr
Homework Helper
To refresh the idea of the remainder theorem, take a look at

http://en.wikipedia.org/wiki/Polynomial_remainder_theorem

In this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?

A cubic polynomial?

rcgldr
Homework Helper
A cubic polynomial?
Ignore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3?

Degree 3?

rcgldr
Homework Helper
Degree 3?
Looking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.

Yup it appears that degree one works as well...

rcgldr
Homework Helper
Yup it appears that degree one works as well...
So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?

ax+b, and I managed to solve it as (2-x).

rcgldr
Homework Helper
To follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0.