Polynom Division: Finding the Reminder with (x-a)(x-b)

[Physicsissuef]

B=A
-B*a+ alpha =-A*b+ beta

Like this?

 

[tiny-tim]

Yes!

So the remainder = px + q = mx + n = … ?

 

[Physicsissuef]

=-A*a+alpha=-A*b+beta
what's next? :) :)

 

[tiny-tim]

=-A*a+alpha=-A*b+beta

Right!

-A*a+alpha=-A*b+beta​

a b alpha and beta were given in the original question.

Your only unknown now is A (now we've got rid of that irritating B )

So A = … ?

 

[Physicsissuef]

$$A= \frac{\alpha - \beta}{a-b}$$

What is next? LOL

 

[tiny-tim]

You're virtually there!

You're just round the corner! (has anyone ever told you that before? …)

$$px + q = A(x-b)\,+\,\beta$$

$$A= \frac{\alpha - \beta}{a-b}$$​

So the remainder = px + q = mx + n = … ?

 

[Physicsissuef]

$$\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}$$
I substitute for B(x-a) + alpha.

 

[tiny-tim]

Hurrah!

Case closed?

 

[Physicsissuef]

Hurrah!

Case closed?

I think so. Thanks buddy

 

[Physicsissuef]

Just, want to ask you, we have:
$$\frac{\beta a-\alpha b}{a-b}$$
and in my textbook result:
$$\frac{\alpha b - \beta a}{a-b}$$

Is their fault?

 

[tiny-tim]

Hi Physicsissuef!

I've gone over it again, and I can't see any mistakes.

Let's test it with a = 2, b = 1, P(x) = $$x^2\,-\,2x\,+\,3$$.

Then alpha = 3, beta = 2.

And (x-a)(x-b) = (x-2)(x-1) = $$x^2\,-\,3x\,+\,2$$, so R(x) = x + 1.

$$\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}$$

= (3-2)x/(2-1) + (2.2 - 3.1)/(2-1)

= x + 1.

So our formula is right, and the textbook is wrong!

Hurrah!