Polynomial Remainder Therem to proove this

[shalikadm]

Homework Statement

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

Homework Equations

the remainder of a polynomial f(x), divided by a linear divisor x-a, is equal to f(a)

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 !

 

[HallsofIvy]

Homework Statement

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

Homework Equations

the remainder of a polynomial f(x), divided by a linear divisor x-a, is equal to f(a)

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

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.

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 !

 

[shalikadm]

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.

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 ?

 

[Curious3141]

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

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

Correct.

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

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.

 

[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).