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Thread: Dividing a quadratic by a linear.

  1. #1
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    Dividing a quadratic by a linear.

    i am lost with problems like these, can someone please help and explain step by step?
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    Last edited by mr fantastic; Feb 25th 2010 at 01:20 AM. Reason: Changed post title
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  2. #2
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    Factorise both the numerator and denominator.

    Numerator:

    $\displaystyle 2x^2-x-3 $

    That can be factorised to: $\displaystyle (x+1)(2x-3) $

    Denominator:

    $\displaystyle x-1 $

    That can be factorised to: $\displaystyle (x+1)(x-1) $

    Now we have a fraction that looks like this:

    $\displaystyle \frac{(x+1)(2x-3)}{(x+1)(x-1)} $

    Now have a think of what you do next....
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  3. #3
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    Quote Originally Posted by jgv115 View Post
    Factorise both the numerator and denominator.

    Numerator:

    $\displaystyle 2x^2-x-3 $

    That can be factorised to: $\displaystyle (x+1)(2x-3) $

    Denominator:

    $\displaystyle x-1 $

    That can be factorised to: $\displaystyle (x+1)(x-1) $
    No, it can't! You were thinking of $\displaystyle x^2- 1$.

    Now we have a fraction that looks like this:

    $\displaystyle \frac{(x+1)(2x-3)}{(x+1)(x-1)} $

    Now have a think of what you do next....
    That should be $\displaystyle \frac{(x+1)(2x-3)}{x-1}$

    By the way, wonderd, you never did say what you wanted to do with that expression!
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  4. #4
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    Quote Originally Posted by wonderd View Post
    i am lost with problems like these, can someone please help and explain step by step?
    You're probably expected to do a polynomial long division to get the result here: simplify (2x^2 - x - 3)/(x - 1) - Wolfram|Alpha

    Your textbook and class notes should have examples ....?
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    Oops Thanks for clearing that up hallsofivy =="
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