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Math Help - Polynomial remainder problem

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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Polynomial remainder problem

    Hello, this is a problem from Israel Gelfand's Algebra book about polynomials.

    Problem 152. The polynomial P gives a remainder of 5x-7 when divided by x^2-1. Find the remainder when P is divided by x-1.

    My try: We know that P=Q(x^2-1)+5x-7 where Q is the quotient, this is the same as P=Q(x+1)(x-1)+5x-7, I see that all I have to do is to rewrite this equation to something like this P=Q(x-1)+5x-7+r, I tried dividing both sides by x+1 but I can't get the form of P=Q(x-1)+R.
    Thanks in advance.
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    Re: Polynomial remainder problem

    \displaystyle \begin{align*} \frac{Q(x + 1)(x - 1) + 5x - 7}{x - 1} &= Q(x + 1) + \frac{5x - 7}{x - 1} \\ &= Q(x + 1) + \frac{5x - 5 - 2}{x + 1} \\ &= Q(x + 1) + \frac{5(x - 1)}{x - 1} - \frac{2}{x - 1} \\ &= Q(x + 1) + 5 - \frac{2}{x - 1} \end{align*}

    So the remainder must be -2.
    Last edited by Prove It; October 23rd 2012 at 07:21 PM.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial remainder problem

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \frac{Q(x + 1)(x - 1) + 5x - 7}{x - 1} &= Q(x + 1) + \frac{5x - 7}{x + 1} \\ &= Q(x + 1) + \frac{5x + 5 - 12}{x + 1} \\ &= Q(x + 1) + \frac{5(x + 1)}{x + 1} - \frac{12}{x + 1} \\ &= Q(x + 1) + 5 - \frac{12}{x + 1} \end{align*}

    So the remainder must be -12.
    Thank you for the reply Prove It, but I have few questions.
    \frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}, isn't this correct? instead of ...=Q(x+1)+\frac{5x-7}{x+1}.
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    Re: Polynomial remainder problem

    Quote Originally Posted by PaulAdrienMauriceDirac View Post
    Thank you for the reply Prove It, but I have few questions.
    \frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}, isn't this correct? instead of ...=Q(x+1)+\frac{5x-7}{x+1}.
    Oops that's a typo. I'll fix it.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial remainder problem

    Quote Originally Posted by Prove It View Post
    Oops that's a typo. I'll fix it.
    It's ok, so should it be something like this?

    \displaystyle \begin{align*}\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}=Q(x+1)+\frac{5x-5-2}{x-1}=Q(x+1)+\frac{5(x-1)}{x-1}-\frac{2}{x-1}=Q(x+1)+5-\frac{2}{x-1}\end{align*}
    and the remainder is -2? I don't understand this, can you explain me why is remainder -2 and not 5-\frac{2}{x-1}? Is it because we divided P by x-1 in the left side?
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    Re: Polynomial remainder problem

    Quote Originally Posted by PaulAdrienMauriceDirac View Post
    It's ok, so should it be something like this?

    \displaystyle \begin{align*}\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}=Q(x+1)+\frac{5x-5-2}{x-1}=Q(x+1)+\frac{5(x-1)}{x-1}-\frac{2}{x-1}=Q(x+1)+5-\frac{2}{x-1}\end{align*}
    and the remainder is -2? I don't understand this, can you explain me why is remainder -2 and not 5-\frac{2}{x-1}? Is it because we divided P by x-1 in the left side?
    The remainder is the numerator of the NON-WHOLE part. 5 is whole.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial remainder problem

    Quote Originally Posted by Prove It View Post
    The remainder is the numerator of the NON-WHOLE part. 5 is whole.
    Hm, can you tell me how did you know that, is that a rule in this type of situations, that remainder is the numerator of the non-whole part?
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    Re: Polynomial remainder problem

    Quote Originally Posted by PaulAdrienMauriceDirac View Post
    Hm, can you tell me how did you know that, is that a rule in this type of situations, that remainder is the numerator of the non-whole part?
    It's a basic rule of writing improper fractions as mixed numbers.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial remainder problem

    Quote Originally Posted by Prove It View Post
    It's a basic rule of writing improper fractions as mixed numbers.
    Yes I know that rule, I was just confused since I learned that P=Q(Divisor)+R, I was confused about which one was quotient and which one was remainder, now I understand, thank you!
    Last edited by PaulAdrienMauriceDirac; October 31st 2012 at 12:34 AM.
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    Re: Polynomial remainder problem

    Remember that remainder is always less than the divisor, this is a little difficult to formalize in case of polynomials rather than in case of numbers. I think, a remainder when divided with a linear polynomial would be a constant, the acceptable remainder when you divide using a quadratic (like x^2 - 1) would be a linear polynomial (5x - 7) or a constant. Keep this in mind, and you will be able to factorize out correctly.

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