polynom division

tiny-tim

Because that equals alpha … can you see why?

And then, what equals beta?

Physicsissuef

but didn't f(x)*(x-a)+px+q=P(x)

So px+q equals alpha. Because alpha is the remainder.

tiny-tim

You've got the principle, but you've got rather confused writing it.

From post #1, P(x) = f(x)(x-a) + alpha.

And so alpha equals pa + q (not px + q), from your post #17, because, as you say, alpha is the remainder.

Right! Nearly there! alpha = pa + q.

Now what does beta equal?

Physicsissuef

pb+q

tiny-tim

oops! my last post was wrong!

should have been:

From post #1, P(x) = f(x)(x-a) + alpha.

Suppose the remainder on dividing f(x) by (x-b) is B.

Then the remainder on diving P(x) by (x-a)(x-b) will be alpha plus B(x-a).

In other words: px + q = alpha plus B(x-a), for some constant B.

Similarly, px + q = beta plus A(x-b), for some constant A.

So alpha plus B(x-a) = beta plus A(x-b)

So …

Sorry!

Physicsissuef

Man, why dividing f(x) with (x-b)?

tiny-tim

Because the remainder on diving P(x) by (x-a)(x-b) is px + q.

So suppose f(x) = g(x)(x-b) + B, for some polynomial g(x), and some constant (remainder) B.

So P(x) = f(x)(x-a) + alpha
= (g(x)(x-b) + B)(x-a) + alpha
= g(x)(x-b)(x-a) + B(x-a) + alpha,
and so the remainder on dividing P(x) by (x-b)(x-a) is B(x-a) + alpha.

So px + q = B(x-a) plus alpha, for some constant B.

Physicsissuef

and why in my text book, the result is
[tex]R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b}[/tex]

??

tiny-tim

That's correct! (except it's [tex]a\beta[/tex], not [tex]\alpha\beta[/tex], at the end)

Physicsissuef

I don't know why.... Please helllppp

tiny-tim

ok, you already have

And similarly you can get

So

So what do B and A equal?

(Remember, you know the result is
[tex]R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b}[/tex];
of course the book is using m and n instead of my p and q.)

Physicsissuef

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

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

What is next?

tiny-tim

Physicsissuef, you're really bad at this!

Don't do any dividing!

Just look at:

You have a polynomial on the left, and another one on the right.

You want these polynomials to be equal (for all values of x).

So …

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

Right!


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 = … ?