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

  • Thread starter Physicsissuef
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What are the two equations involving (px + q) and alpha and beta?px + q = alpha plus B(x-a) … (1)px + q = beta plus A(x-b) … (2)Subtract (1) from (2) … what do you get?
  • #36
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:
 
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  • #37
[tex]\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}[/tex]
I substitute for B(x-a) + alpha.
 
  • #38
Hurrah!

Case closed? :smile:
 
  • #39
tiny-tim said:
Hurrah!

Case closed? :smile:

I think so. Thanks buddy :smile:
 
  • #40
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?
 
  • #41
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:
 

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