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Math Help - remainder of p(x)/g(x) where g(x) is quadratic

  1. #1
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    remainder of p(x)/g(x) where g(x) is quadratic

    The chapter is about Polynomials and the there are few questions on remainder theorem. If p(x) is divided by (ax-b) where a and b are constants, the remainder will be b/a. The teacher gave one extra out-of-portion question.

    Find the remainder when $x^{2014}+x^{2013}$ is divided by $x^2-1$.

    The divisor is quadratic. So the remainder will either be linear or constant. I don't know how to find remainder here. But I found it using division $x+1$. Is there any way to find the remainder here
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    Quote Originally Posted by NameIsHidden View Post
    If p(x) is divided by (ax-b) where a and b are constants, the remainder will be b/a.
    You must mean p(b/a).

    Quote Originally Posted by NameIsHidden View Post
    Find the remainder when $x^{2014}+x^{2013}$ is divided by $x^2-1$.
    Let $p(x)=q(x)(x^2-1)+ax+b$ for some $a,b$. What happens when you apply the idea behind the proof of the fact you quoted above (the polynomial remainder theorem)? The idea there is to instantiate $x$ with the root of the divisor ($x^2-1$ in this case).
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    Quote Originally Posted by emakarov View Post
    You must mean p(b/a).
    Sorry a mistake in a hurry.

    Quote Originally Posted by emakarov View Post
    Let $p(x)=q(x)(x^2-1)+ax+b$ for some $a,b$. What happens when you apply the idea behind the proof of the fact you quoted above (the polynomial remainder theorem)? The idea there is to instantiate $x$ with the root of the divisor ($x^2-1$ in this case).
    The roots of $x^2-1$ are $\pm 1$.

    Let $x= \pm 1$.

    Then
    $p(1)=a+b=2$

    $p(-1)=-a+b=0$(by evaluating p(-1) and p(1))

    Is there any way to figure out $a$ and $b$?
    Last edited by NameIsHidden; June 24th 2014 at 08:25 AM.
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    Quote Originally Posted by NameIsHidden View Post
    Then
    $p(1)=a+b$

    $p(-1)=-a+b$

    Is there any way to figure out $a$ and $b$?
    Yes, if you know how to solve a system of two linear equations with two unknowns...
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    The remainder should be linear. Shouldn't it? If I substitute two equations, I will get a constant. Can you explain a little more?
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    Quote Originally Posted by NameIsHidden View Post
    The remainder should be linear. Shouldn't it?
    Not necessarily, though in this case it is. The restriction on the remainder is that its degree must me smaller than that of the divisor, which is 2. Therefore, the degree of the remainder can be either 1 or 0.

    Quote Originally Posted by NameIsHidden View Post
    If I substitute two equations, I will get a constant.
    I don't agree. Please check if the $a$ and $b$ you found satisfy the two equations.
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    $p(1)=1^{2014}+1^{2013}=1+1=2=a+b$ Is the remainder two here?
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    No, the remainder is $ax+b$ where $a$ and $b$ are the solutions of the system
    \[
    \left\{\begin{aligned} a+b&=2\\ -a+b&=0\end{aligned}\right.
    \]
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    Thank you very much. I solved it, $a=b=1$. Thus $r(x)=1x+1=x+1$
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    Re: remainder of p(x)/g(x) where g(x) is quadratic

    \dfrac{x^{2014}+x^{2013}}{x^2-1} = \dfrac{x^{2013}(x+1)}{(x+1)(x-1)} = \dfrac{x^{2013}}{x-1} = \dfrac{x^{2013}-1}{x-1}+\dfrac{1}{x-1}

    Does that help at all?

    Seems like the remainder is x+1, just as you found above.
    Last edited by SlipEternal; June 25th 2014 at 07:10 AM.
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