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Math Help - (r-s)(x)

  1. #1
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    (r-s)(x)

    Suppose r(x) = 3+4/x^2+1, s(x)= x^+6/6x-1. Write the expression (r-s)(x) as a simplified ratio with the numerator and denominator each written as a sum of terms of the form cx^m and c>0 for the term with the highest power in the numerator. I ended up with (-x^4+11x^2+21x-10)/-6x^3+x^2-6x+1. Where did I go wrong here?
    Last edited by Eraser147; October 2nd 2012 at 03:20 PM.
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  2. #2
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    Re: (r-s)(x)

    Without knowing the supposition, it is difficult to say where you went wrong.
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    Re: (r-s)(x)

    This is the right answer (r-s)(x)-session.quest1115433entrance1_n10164.gif but I don't know how to get it.
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  4. #4
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    Re: (r-s)(x)

    The links to images in your first post are broken for me.
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    Re: (r-s)(x)

    Edited.
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  6. #6
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    Re: (r-s)(x)

    Quote Originally Posted by Eraser147 View Post
    Suppose r(x) = 3+4/x^2+1, s(x)= x^+6/6x-1.
    Why not add 3 + 1 and simplify r(x) to 4 + (4/x^2)? And why do you write + in front of 6 in x^+6?
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    Re: (r-s)(x)

    The only way I got the answer right was when I subtracted s with r instead of r - s. It looked like this with the polynomial subtraction (x^4 + 7x^2 + 6)-(18x^2+21x-4). And it gives me x^4-11x^2+21x+10 for the numerator.
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    Re: (r-s)(x)

    Despite three request to write the question correctly, I am still doubting this has been done.
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  9. #9
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    Re: (r-s)(x)

    (r-s)(x)-problem.jpgHere. it has to be correct. Just click on the image to enlarge.
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    Re: (r-s)(x)

    \begin{align*}\frac{3x+4}{x^2+1}-\frac{x^2+6}{6x-1}&=\frac{(6x-1)(3x+4)-(x^2+1)(x^2+6)}{(x^2+1)(6x-1)}\\&=\frac{18x^2+21x-4-(x^4+7x^2+6)}{6x^3-x^2+6x-1}\\&=\frac{-x^4+11x^2+21x-10}{6x^3-x^2+6x-1}\\&=\frac{x^4-11x^2-21x+10}{-6x^3+x^2-6x+1}\end{align*}
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  11. #11
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    Re: (r-s)(x)

    Why are the signs flipped again at the end?
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  12. #12
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    Re: (r-s)(x)

    Changing the sign of both the numerator and the denominator (i.e., multiplying both by -1) does not change the fraction. The problem asks for a fraction with a positive leading coefficient in the numerator.
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    Re: (r-s)(x)

    Quote Originally Posted by Eraser147 View Post
    Suppose r(x) = 3+4/x^2+1,

    Do you mean 3+ 4/(x^2+ 1)?

    s(x)= x^+6/6x-1.

    What should be after the "^"? Do you mean x^?+ 6/(6x- )? Or (x^?+ 6)(6x- 1)?

    Write the expression (r-s)(x) as a simplified ratio with the numerator and denominator each written as a sum of terms of the form cx^m and c>0 for the term with the highest power in the numerator. I ended up with (-x^4+11x^2+21x-10)/-6x^3+x^2-6x+1. Where did I go wrong here?
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  14. #14
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    Re: (r-s)(x)

    Oh, I already figured it out. Emakarov showed it. My mistake was not flipping the signs by multiplying -1 from numerator and denominator since the problem asked for the leading term to be a positive number and the constant to be more than zero. Read the post above to find the image of the actual equation if you are interested.
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