Finding the remainder when dividing a polynomial by (x-a)(x-b)

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

Like this?
 
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Yes! :smile: :smile:

So the remainder = px + q = mx + n = … ? :smile:
 
=-A*a+alpha=-A*b+beta
what's next? :) :)
 
Physicsissuef said:
=-A*a+alpha=-A*b+beta

Right! :smile: :smile:

-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 :mad:)

So A = … ? :smile:
 
[tex]A= \frac{\alpha - \beta}{a-b}[/tex]

What is next? LOL :smile:
 
You're virtually there!

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

[tex]px + q = A(x-b)\,+\,\beta[/tex]

[tex]A= \frac{\alpha - \beta}{a-b}[/tex]​

So the remainder = px + q = mx + n = … ? :smile:
 
[tex]\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}[/tex]
I substitute for B(x-a) + alpha.
 
tiny-tim said:
Hurrah!

Case closed? :smile:

I think so. Thanks buddy :smile:
 
Just, want to ask you, we have:
[tex]\frac{\beta a-\alpha b}{a-b}[/tex]
and in my textbook result:
[tex]\frac{\alpha b - \beta a}{a-b}[/tex]

Is their fault?
 
Hi Physicsissuef!:smile:

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

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

Then alpha = 3, beta = 2.

And (x-a)(x-b) = (x-2)(x-1) = [tex]x^2\,-\,3x\,+\,2[/tex], so R(x) = x + 1.

[tex]\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}[/tex]

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

= x + 1.

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

Hurrah! :biggrin: