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Math Help - factoring help

  1. #1
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    factoring help

    I'm struggling with factoring. Some are really easy, and some I don't know where to start.

    m3 -mn2 +m2n - n3 + m2 - n2 is one that I need to factor.

    The answer is (m+n)(m-n)(m+n+1) and if I multiply it all out, it gets me back to the original.

    I get that (m+n)(m-n) is (m2 - n2). Do the substitution and you get m3 -mn2 +m2n - n3 + (m+n)(m-n).

    You can factor m3 -mn2 +m2n - n3 into (m2-n2)(m+n), and further into (m+n)(m-n)(m+n).

    Substitute again and you get (m+n)(m-n)(m+n) + (m+n)(m-n). Then I don't know where to go.

    Can someone point me in the right direction?

    Thanks
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  2. #2
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    Re: factoring help

    Quote Originally Posted by baldysm View Post
    Substitute again and you get (m+n)(m-n)(m+n) + (m+n)(m-n). Then I don't know where to go.
    You're so close! Just factorise the (m+n)(m-n)

    (m+n)(m-n)(m+n) + (m+n)(m-n)=(m+n)(m-n)\bigg((m+n)+1\bigg)
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  3. #3
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    Re: factoring help

    Hmmm, lost me.

    Are you factoring the (m+n)(m-n) from (m+n)(m-n)(m+n)?

    Guess I don't see how to factor (m+n)(m-n) into (m+n+1)
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  4. #4
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    Re: factoring help

    Quote Originally Posted by baldysm View Post
    Hmmm, lost me.
    Are you factoring the (m+n)(m-n) from (m+n)(m-n)(m+n)?
    No, I'm factoring  (m+n)(m-n) from the whole expression (m+n)(m-n)(m+n) + (m+n)(m-n) .

    I suppose it is easier to see if you use variables: Let y=(m+n)(m-n). Then your expression becomes y(m+n)+1y. Factorising y gives y((m+n)+1)=(m+n)(m-n)(m+n+1)
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  5. #5
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    Re: factoring help

    Ahhh, got it! Makes sense now.

    Thanks!
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  6. #6
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    Re: factoring help

    Hello, baldysm!

    I will change the variables.


    \text{Factor: }\:a^3 - ab^2 + a^2b - b^3 + a^2-b^2

    We have: . (a^3-ab^2) + (a^2b - b^3) + (a^2-b^2)

    Factor: . a(a^2-b^2) + b(a^2-b^2) + (a^2-b^2)

    Factor out (a^2-b^2)\!:\;\;(a^2-b^2)(a + b + 1)

    Therefore: . (a-b)(a+b)(a+b+1)
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