# A tricky remainder theorem problem

• sooyong94
In summary, the equation for the polynomial Q(x)(x-1)(x+1) - x + 2 is (2-x).So the equation for the polynomial Q(x)(x-1)(x+1) - x + 2 is (2-x).

## 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.

sooyong94 said:
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".

sooyong94 said:
explanation

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

sooyong94 said:
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?

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?

sooyong94 said:
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?

sooyong94 said:
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...

sooyong94 said:
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).

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.