- #31
Physicsissuef
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B=A
-B*a+ alpha =-A*b+ beta
Like this?
-B*a+ alpha =-A*b+ beta
Like this?
Polynom Division: Finding the Reminder with (x-a)(x-b)
- Thread starter Physicsissuef
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Homework Help Overview
The discussion revolves around polynomial division, specifically finding the remainder when dividing a polynomial P(x) by the product of two linear factors, (x-a)(x-b). Participants explore the relationship between the remainders obtained from dividing P(x) by x-a and x-b individually.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the form of the polynomial P(x) and the nature of the remainder R(x) when divided by (x-a)(x-b). There are attempts to express R(x) in terms of constants p and q, and to derive equations based on the known remainders alpha and beta. Questions arise regarding the assumptions about the degree of the polynomial and the implications of polynomial division rules.
Discussion Status
The discussion is active, with participants providing hints and guidance on how to express the remainder in terms of the known values. There is an ongoing exploration of the relationships between the constants involved and the remainders from previous divisions. Multiple interpretations and approaches are being considered, indicating a productive exchange of ideas.
Contextual Notes
Participants note the constraints of polynomial division, particularly that the remainder must be of lower degree than the divisor. There are references to specific values for a and b, and how these affect the calculations and results discussed.
- #32
tiny-tim
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Yes! 
So the remainder = px + q = mx + n = … ?
So the remainder = px + q = mx + n = … ?
- #33
Physicsissuef
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=-A*a+alpha=-A*b+beta
what's next? :) :)
what's next? :) :)
- #34
tiny-tim
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Physicsissuef said:
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 = … ?
- #35
Physicsissuef
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A= \frac{\alpha - \beta}{a-b}
What is next? LOL
What is next? LOL
- #36
tiny-tim
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You're virtually there!
You're just round the corner! (has anyone ever told you that before? …)
So the remainder = px + q = mx + n = … ?
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}
A= \frac{\alpha - \beta}{a-b}
So the remainder = px + q = mx + n = … ?
- #37
Physicsissuef
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\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}
I substitute for B(x-a) + alpha.
I substitute for B(x-a) + alpha.
- #38
tiny-tim
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Hurrah!
Case closed?
Case closed?
- #39
Physicsissuef
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tiny-tim said:Hurrah!
Case closed?
I think so. Thanks buddy
- #40
Physicsissuef
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Just, want to ask you, we have:
\frac{\beta a-\alpha b}{a-b}
and in my textbook result:
\frac{\alpha b - \beta a}{a-b}
Is their fault?
\frac{\beta a-\alpha b}{a-b}
and in my textbook result:
\frac{\alpha b - \beta a}{a-b}
Is their fault?
- #41
tiny-tim
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Hi Physicsissuef!
I've gone over it again, and I can't see any mistakes.
Let's test it with a = 2, b = 1, P(x) = x^2\,-\,2x\,+\,3.
Then alpha = 3, beta = 2.
And (x-a)(x-b) = (x-2)(x-1) = x^2\,-\,3x\,+\,2, so R(x) = x + 1.
\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}
= (3-2)x/(2-1) + (2.2 - 3.1)/(2-1)
= x + 1.
So our formula is right, and the textbook is wrong!
Hurrah!
I've gone over it again, and I can't see any mistakes.
Let's test it with a = 2, b = 1, P(x) = x^2\,-\,2x\,+\,3.
Then alpha = 3, beta = 2.
And (x-a)(x-b) = (x-2)(x-1) = x^2\,-\,3x\,+\,2, so R(x) = x + 1.
\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}
= (3-2)x/(2-1) + (2.2 - 3.1)/(2-1)
= x + 1.
So our formula is right, and the textbook is wrong!
Hurrah!
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