# Polynom division

by Physicsissuef
Tags: division, polynom
 P: 908 but didn't f(x)*(x-a)+px+q=P(x) So px+q equals alpha. Because alpha is the remainder.
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Thanks
P: 26,148
 Quote by Physicsissuef but didn't f(x)*(x-a)+px+q=P(x) So px+q equals alpha. Because alpha is the remainder.
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?
 P: 908 pb+q
 Sci Advisor HW Helper Thanks P: 26,148 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!
 P: 908 Man, why dividing f(x) with (x-b)?
 Sci Advisor HW Helper Thanks P: 26,148 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.
 P: 908 and why in my text book, the result is $$R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b}$$ ??
 Sci Advisor HW Helper Thanks P: 26,148 That's correct! (except it's $$a\beta$$, not $$\alpha\beta$$, at the end) You tell me why!
P: 908
 Quote by tiny-tim That's correct! (except it's $$a\beta$$, not $$\alpha\beta$$, at the end) You tell me why!
I don't know why.... Please helllppp
 Sci Advisor HW Helper Thanks P: 26,148 ok, you already have $$px + q = B(x-a)\,+\,\alpha\,.$$ And similarly you can get $$px + q = A(x-b)\,+\,\beta\,.$$ So $$B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\beta\,.$$ So what do B and A equal? (Remember, you know the result is $$R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b}$$; of course the book is using m and n instead of my p and q.)
 P: 908 $$B= \frac{A(x-b)}{x-a}+ \frac{\beta - \alpha}{x-a}$$ $$A= \frac{B(x-a)}{x-b}+ \frac{\alpha - \beta}{x-b}$$ What is next?
 Sci Advisor HW Helper Thanks P: 26,148 Physicsissuef, you're really bad at this! Don't do any dividing! Just look at: $$B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\beta\,.$$ 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 …
 P: 908 B=A -B*a+ alpha =-A*b+ beta Like this?
 Sci Advisor HW Helper Thanks P: 26,148 Yes!! So the remainder = px + q = mx + n = … ?
 P: 908 =-A*a+alpha=-A*b+beta what's next? :) :)
HW Helper
Thanks
P: 26,148
 Quote by Physicsissuef =-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 = … ?
 P: 908 $$A= \frac{\alpha - \beta}{a-b}$$ What is next? LOL
 Sci Advisor HW Helper Thanks P: 26,148 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 = … ?

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