
#19
Mar1008, 05:40 PM

P: 909

but didn't f(x)*(xa)+px+q=P(x)
So px+q equals alpha. Because alpha is the remainder. 



#20
Mar1008, 05:58 PM

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From post #1, P(x) = f(x)(xa) + 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? 



#21
Mar1108, 02:41 AM

P: 909

pb+q




#22
Mar1108, 03:12 AM

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oops! my last post was wrong!
should have been: From post #1, P(x) = f(x)(xa) + alpha.Sorry! 



#23
Mar1108, 03:37 AM

P: 909

Man, why dividing f(x) with (xb)?




#24
Mar1108, 03:48 AM

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Because the remainder on diving P(x) by (xa)(xb) is px + q.
So suppose f(x) = g(x)(xb) + B, for some polynomial g(x), and some constant (remainder) B. So P(x) = f(x)(xa) + alpha = (g(x)(xb) + B)(xa) + alpha = g(x)(xb)(xa) + B(xa) + alpha, and so the remainder on dividing P(x) by (xb)(xa) is B(xa) + alpha. So px + q = B(xa) plus alpha, for some constant B. 



#25
Mar1108, 04:02 AM

P: 909

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



#26
Mar1108, 05:29 AM

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




#27
Mar1108, 05:39 AM

P: 909





#28
Mar1108, 05:53 AM

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ok, you already have
[tex]px + q = B(xa)\,+\,\alpha\,.[/tex] And similarly you can get [tex]px + q = A(xb)\,+\,\beta\,.[/tex] So [tex]B(xa)\,+\,\alpha\,=\,A(xb)\,+\,\beta\,.[/tex] So what do B and A equal? (Remember, you know the result is [tex]R(x)=mx+n= \frac{\alpha  \beta}{ab}x + \frac{b\alpha  \alpha\beta}{ab}[/tex]; of course the book is using m and n instead of my p and q.) 



#29
Mar1108, 06:04 AM

P: 909

[tex]B= \frac{A(xb)}{xa}+ \frac{\beta  \alpha}{xa}[/tex]
[tex]A= \frac{B(xa)}{xb}+ \frac{\alpha  \beta}{xb}[/tex] What is next? 



#30
Mar1108, 06:18 AM

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Physicsissuef, you're really bad at this!
Don't do any dividing! Just look at: [tex]B(xa)\,+\,\alpha\,=\,A(xb)\,+\,\beta\,.[/tex] 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 … 



#31
Mar1108, 06:28 AM

P: 909

B=A
B*a+ alpha =A*b+ beta Like this? 



#32
Mar1108, 06:39 AM

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



#33
Mar1108, 06:41 AM

P: 909

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



#34
Mar1108, 06:53 AM

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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 = … ? 



#35
Mar1108, 06:57 AM

P: 909

[tex]A= \frac{\alpha  \beta}{ab}[/tex]
What is next? LOL 


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