Remainder of polynomial division

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Discussion Overview

The discussion revolves around the properties of polynomial division, specifically regarding the remainders when dividing a polynomial by another polynomial. Participants explore whether a specific relationship holds true when raising the polynomial to a power.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants question whether the remainder of f(x) divided by g(x) being a constant a implies that the remainder of (f(x))^n divided by g(x) is also a constant.
  • One participant suggests starting with the definition of remainders in polynomial division, indicating that f(x) can be expressed as q(x)·g(x) + a.
  • Another participant proposes using the binomial theorem to explore the relationship further by raising both sides of the equation to the power of n.
  • A later reply indicates that after testing the hypothesis, they found it to be correct.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial claim, as some express uncertainty while others confirm the relationship through examples. The discussion remains partially unresolved with differing viewpoints on the validity of the claim.

Contextual Notes

Limitations include the lack of formal proof or rigorous argumentation for the proposed relationship, as well as dependence on specific examples that may not generalize.

songoku
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Is this true?

If the remainder of f(x) / g(x) is a (where a is constant), then the remainder of (f(x))n / g(x) is an

I don't know how to be sure whether it is correct or wrong. I just did several examples and it works.

Thanks
 
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songoku said:
Is this true?

If the remainder of f(x) / g(x) is a (where a is constant), then the remainder of (f(x))n / g(x) is an

I don't know how to be sure whether it is correct or wrong. I just did several examples and it works.

Thanks
One way to deal with remainders is to write down what it means, which is always a good point to start at.
Here you have ##f(x)=q(x)\cdot g(x) + a##. Now you can work with it.
 
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I tried it and found out it is actually correct.

Thanks a lot for the help
 

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