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Math Help - Factorising

  1. #1
    Member princess_anna57's Avatar
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    Smile Factorising

    solve (x-5)(x+2) = 3x+2

    Here's what I got so far:

    x^2 + 2x - 5x - 10 = 3x + 2
    x^2 - 3x - 10 = 3x + 2
    x^2 - 6x - 12 = 0

    and then I don't know what to do next. Do you do (x ) (x )? I don't know how to figure out how to get the missing numbers.

    I would also like to know how to solve these two:
    5x = -2x^3
    5x = 2x^3
    Last edited by princess_anna57; July 11th 2008 at 05:30 PM. Reason: more to add
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Smile

    Quote Originally Posted by princess_anna57 View Post
    solve (x-5)(x+2) = 3x+2

    Here's what I got so far:

    x^2 + 2x - 5x - 10 = 3x + 2
    x^2 - 3x - 10 = 3x + 2
    x^2 - 6x - 12 = 0

    and then I don't know what to do next. Do you do (x ) (x )? I don't know how to figure out how to get the missing numbers.
    Complete the square or use the quadratic formula. You won't get nice factors with integer coefficients for this one.

    Quote Originally Posted by princess_anna57 View Post
    I would also like to know how to solve these two:
    5x = -2x^3
    5x = 2x^3
    This one should be easy.

    5x = -2x^3\Rightarrow 2x^3 + 5x = 0\Rightarrow x(2x^2 + 5) = 0

    Only one of the solutions will be real.
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