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Math Help - is this correct..

  1. #1
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    is this correct..

    Factoring the following;

    x^3-9x=0

    = x(x^2-9)=0

    = (x-3)(x+3)=0

    = x=3 x=-3


    Thnaks
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  2. #2
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    Quote Originally Posted by Morzilla View Post
    Factoring the following;
    x^3-9x=0
    = x(x^2-9)=0
    = (x-3)(x+3)=0
    = x=3 x=-3
    What happened to x=0?
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  3. #3
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    Quote Originally Posted by Plato View Post
    What happened to x=0?
    so are you saying that the x in x(x-3)(x+3) should also be accounted for?
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  4. #4
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    Quote Originally Posted by Morzilla View Post
    so are you saying that the x in x(x-3)(x+3) should also be accounted for?
    Well of course!
    If you had \left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - \pi } \right) = 0 then the solutions would be
    x =  - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi .
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  5. #5
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    Quote Originally Posted by Plato View Post
    Well of course!
    If you had \left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - \pi } \right) = 0 then the solutions would be
    x =  - 1\,,\,x = 0\,,\,x = 1\,,\,x = 3\,,\,\,\& \,x = \pi .
    ......ohhhhhhh! I see, I see ok thank you very much!

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