A tricky remainder theorem problem

[sooyong94]

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.

 

[Mastermind01]

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.

 

[sooyong94]

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

 

[Ray Vickson]

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]

explanation

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

 

[Ray Vickson]

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

Go back and read post #2 again, then read your question in post #3.

 

[sooyong94]

So it has something to do with the divisor right?

 

[rcgldr]

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?

 

[sooyong94]

A cubic polynomial?

 

[rcgldr]

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?

 

[sooyong94]

Degree 3?

 

[rcgldr]

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.

 

[sooyong94]

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

 

[rcgldr]

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) ?

 

[sooyong94]

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

 

[rcgldr]

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.