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Thread: Thinking Question Help plz

  1. #1
    Bvs
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    Thinking Question Help plz

    Hi guys, so i need a plan to execute the steps and follow these criterias but i cant even get a head start. Any help would be much appreciated. Thanks guys!

    Two rational variable expressions are divided and the simplified expression must meet the following conditions:

    —The simplified expression's denominator is a linear expression while the numerator is a quadratic expression
    —The simplified expression's numerator must contain a factor of (3�� − 5)
    —The restrictions for the expressions must include �� ≠ −2, 3
    Last edited by Bvs; Apr 19th 2018 at 03:12 AM.
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  2. #2
    MHF Contributor

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    Re: Thinking Question Help plz

    Quote Originally Posted by Bvs View Post
    Hi guys, so i need a plan to execute the steps and follow these criterias but i cant even get a head start. Any help would be much appreciated. Thanks guys!

    Two rational variable expressions are divided and the simplified expression must meet the following conditions:

    —The simplified expression's denominator is a linear expression while the numerator is a quadratic expression
    Do know what these words mean? Since we have "denominator" and "numerator" we must have a fraction. A "linear expression" is of the form "ax+ b" and a "quadratic expression" is of the form "cx^2+ dx+ e" where a, b, c, d, and e are specific numbers.

    —The simplified expression's numerator must contain a factor of (3�� − 5)
    You seem to have a symbol here that does not show on my computer. I will assume that was simply "3x- 5". Since the numerator must a 'quadratic expression' with that factor, it must be of the form (3x- 5)(px+ q).

    —The restrictions for the expressions must include �� ≠ −2, 3
    Now this one puzzles me. The only "restriction" for a fraction is that the denominator cannot be 0. But since the denominator is to be a "linear expression", of the form ax+ b, there is only one number, -b/a, that would make the denominator 0, not two.
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  3. #3
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    Re: Thinking Question Help plz

    $\displaystyle \dfrac{(3x-5)(x-3)(x+4)}{(x-3)(x+2)}$ simplifies to $\displaystyle \dfrac{3x^2+7x-20}{x+2}$ when $\displaystyle x \neq 3$.
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