# Homework Help: Polynomial Remainder Therem to proove this

1. Jul 25, 2012

1. The problem statement, all variables and given/known data
Applying remainder theorem again and again to show that the remainder of the f(x) polynomial function when divided by (x-α)(x-β) is A(x-α)+B . Determine A and B
2. Relevant equations
the remainder of a polynomial f(x), divided by a linear divisor x-a, is equal to f(a)
3. The attempt at a solution
Ok...here's how the teacher has solve this...

f(x)=(x-α)g(x)+B(remainder theorem)
g(x)=(x-β)∅(x)+A(remainder theorem)

f(x)=(x-α)[(x-β)∅(x)+A]+B
f(x)=(x-α)(x-β)∅(x)+A(x-α)+B
∅(x)$\rightarrow$quotient
A(x-α)+B$\rightarrow$remainder

But I think that he has forgotten to use the remainder theorem there..I can't see where he has applied it..

I think if we use the theorem,we have to do something like this.

f(x)=(x-α)g(x)+α
g(x)=(x-β)∅(x)+β

f(x)=(x-α)[(x-β)∅(x)+β]+σ
f(x)=(x-α)(x-β)∅(x)+β(x-α)+σ

Here we get something like A(x-α)+B as the remainder{β(x-α)+σ}..But I think that I'm wrong as its too easy then to determine A and B...like A=β and B=α...

Please someone show me were has he used the remainder theorem..Thanks !

2. Jul 25, 2012

### HallsofIvy

You are using the remainder theorem incorrectly here. At x= α the remainder of f(x) is f(α), which your instructor called "A", not α. Similarly, at x= β, the remander of g(x) is g(β), which your instructor called "B", not β.

3. Jul 25, 2012

Oh...sorry..I have written it wrong...

f(x)=(x-α)g(x)+f(α)(remainder theorem)
g(x)=(x-β)∅(x)+f(β)(remainder theorem)

f(x)=(x-α)[(x-β)∅(x)+β]+f(σ)
f(x)=(x-α)(x-β)∅(x)+f(β)(x-α)+f(σ)

Here we get something like A(x-α)+B as the remainder{f(β)(x-α)+f(σ)}
A=f(β) and B=f(σ)

Correct now ?
now where I have gone wrong ?

4. Aug 4, 2012

### Curious3141

Correct.

Mistake here. Shouldn't that remainder be g(β)?

Also, for some strange reason, you started using sigma (σ) instead of alpha (α) later in your post.

5. Aug 4, 2012

### Curious3141

You will need to express g(β) purely in terms of f(β), f(α), α and β after that (which is what is involved in determining "A"). You may use the first equation to do this.

There is a subtlety here. This method works fine when β is not equal to α, but fails when β = α (why?). The question did not stipulate inequality, so you'll need to cover both cases.

In the case where β = α, you'll need another trick to find g(α). This involves a little calculus (differentiation).