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Math Help - Rtional Expression...re-submit

  1. #1
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    Rtional Expression...re-submit

    A student was given the following poblem:

    4x^2 - 9
    ___________ * 6x - 9

    4x^ + 12 +9

    He states his answer to the problem as 3(2x - 3). Is the answer correct? If not, explain why. What would you suggest he do to solve the problem.


    The above is the problem as presented to me. I do not have a clue as to how to answer it! Thanks
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  2. #2
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    Quote Originally Posted by Isdene View Post
    A student was given the following poblem:

    4x^2 - 9
    ___________ * 6x - 9

    4x^ + 12 +9

    Mr F says: I assume that the denominator is 4x^2 + 12x + 9. Although, speculating on the intent of the question, I'll bet it's meant to be 4x^2 - 12x + 9 ......

    He states his answer to the problem as 3(2x - 3). Is the answer correct? If not, explain why. What would you suggest he do to solve the problem.


    The above is the problem as presented to me. I do not have a clue as to how to answer it! Thanks
    If you factorise you get:

    \frac{(2x - 3)(2x + 3)3(2x - 3)}{(2x + 3)^2}.

    You can cancel the (2x + 3) as a common factor provided 2x + 3 \neq 0 \Rightarrow x \neq -\frac{3}{2}. Otherwise you're dividing by zero and that's against the rules (on the grounds that everything would be totally insane if you did).

    So the answer is \frac{3(2x - 3)^2}{2x + 3}, x \neq -\frac{3}{2}.

    So the student is wrong on two counts:

    - they don't have the x \neq -\frac{3}{2} bit.

    - they have 3(2x - 3) instead of \frac{3(2x - 3)^2}{2x + 3}.

    So to solve the problem I'd suggest the student get urgent remedial mathematics help.
    Last edited by mr fantastic; February 15th 2008 at 06:35 PM. Reason: Mucked up the factorising because I was looking at the students bogus answer. And decided to be a wiseguy - see last sentance
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