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Math Help - Factorsing by removing a common factor

  1. #1
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    Factorsing by removing a common factor

    I can factorize problems like this:
    3x^2 + 5x
    3x^2: (1) (3) (x) (x)
    5x: (1) (5) (x)
    Using 1 and x to drop outside and use the leftovers:
    x(3x + 5) *final*

    But I'm being stumped by this kind of problem:
    (x + 2)^2 - 5(x + 2) ?
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by robbz View Post
    But I'm being stumped by this kind of problem:
    (x + 2)^2 - 5(x + 2) ?
    First expand both sets of brackets:

    (x + 2)^2 - 5(x + 2) = x^2+4x+4-5x-10

    Collect like terms:

    x^2-x-6

    Then factorise like you would a normal quadratic equation (I am assuming that you have come across these before, if not please say).

    Hope this helps.
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  3. #3
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    Ok, so when you left off at x^2 - x - 6.

    How do I factorize that? It's longer than the binomials I've been doing.
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  4. #4
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    Think about it like this:

    <br />
(x + 2)^2 - 5(x + 2) ~=~ (x+2)\textcolor{red}{(x+2)} - (5)\textcolor{red}{(x+2)}<br />

    So, we see that (x+2) is common in both terms, so we factor the whole thing out:

    \textcolor{red}{(x+2)} \left[ (x+2) - 5 \right]

    Then you can simplify the right half.
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  5. #5
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    Robbz, drumist's response is the one to go with!
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  6. #6
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    Thanks for the help everyone.
    I understand it now.
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