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Polynom division

by Physicsissuef
Tags: division, polynom
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Physicsissuef
#19
Mar10-08, 05:40 PM
P: 910
but didn't f(x)*(x-a)+px+q=P(x)

So px+q equals alpha. Because alpha is the remainder.
tiny-tim
#20
Mar10-08, 05:58 PM
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Quote Quote by Physicsissuef View Post
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?
Physicsissuef
#21
Mar11-08, 02:41 AM
P: 910
pb+q
tiny-tim
#22
Mar11-08, 03:12 AM
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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
#23
Mar11-08, 03:37 AM
P: 910
Man, why dividing f(x) with (x-b)?
tiny-tim
#24
Mar11-08, 03:48 AM
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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
#25
Mar11-08, 04:02 AM
P: 910
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
#26
Mar11-08, 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!
Physicsissuef
#27
Mar11-08, 05:39 AM
P: 910
Quote Quote by tiny-tim View Post
That's correct! (except it's [tex]a\beta[/tex], not [tex]\alpha\beta[/tex], at the end)

You tell me why!
I don't know why.... Please helllppp
tiny-tim
#28
Mar11-08, 05:53 AM
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ok, you already have
[tex]px + q = B(x-a)\,+\,\alpha\,.[/tex]

And similarly you can get
[tex]px + q = A(x-b)\,+\,\beta\,.[/tex]

So
[tex]B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\beta\,.[/tex]

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
#29
Mar11-08, 06:04 AM
P: 910
[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
#30
Mar11-08, 06:18 AM
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Physicsissuef, you're really bad at this!

Don't do any dividing!

Just look at:
[tex]B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\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
Physicsissuef
#31
Mar11-08, 06:28 AM
P: 910
B=A
-B*a+ alpha =-A*b+ beta

Like this?
tiny-tim
#32
Mar11-08, 06:39 AM
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Yes!!

So the remainder = px + q = mx + n = ?
Physicsissuef
#33
Mar11-08, 06:41 AM
P: 910
=-A*a+alpha=-A*b+beta
what's next? :) :)
tiny-tim
#34
Mar11-08, 06:53 AM
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Quote Quote by Physicsissuef View Post
=-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 = ?
Physicsissuef
#35
Mar11-08, 06:57 AM
P: 910
[tex]A= \frac{\alpha - \beta}{a-b}[/tex]

What is next? LOL
tiny-tim
#36
Mar11-08, 07:08 AM
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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 = ?


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