# A tricky remainder theorem problem

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1. Jun 12, 2016

### sooyong94

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
Remainder theorem

3. 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.

2. Jun 12, 2016

### 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.

3. Jun 13, 2016

### sooyong94

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

4. Jun 13, 2016

### Ray Vickson

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".

5. Jun 13, 2016

### sooyong94

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

6. Jun 13, 2016

### Ray Vickson

7. Jun 13, 2016

### sooyong94

So it has something to do with the divisor right?

8. Jun 13, 2016

### 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?

9. Jun 13, 2016

### sooyong94

A cubic polynomial?

10. Jun 13, 2016

### rcgldr

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?

11. Jun 13, 2016

### sooyong94

Degree 3?

12. Jun 13, 2016

### rcgldr

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.

13. Jun 13, 2016

### sooyong94

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

14. Jun 13, 2016

### rcgldr

So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?

15. Jun 13, 2016

### sooyong94

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

16. Jun 13, 2016

### 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.