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Math Help - Factoring and the Distributive Property Question

  1. #1
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    Factoring and the Distributive Property Question

    Hi all,

    I've been struggling a little with my algebra so during the summer break I've been working through the book Algebra Demystified and it's really helping.

    But I've come across a question half way through that I can't solve, I'm not sure how to tackle it, the question and answer are

    x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)

    Any tips or a starting point would be greatly appreciated.

    Thanks.
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  2. #2
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    Re: Factoring and the Distributive Property Question

    Quote Originally Posted by Srengam View Post
    the question and answer are

    x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)
    This is not a question; this is an equality.
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  3. #3
    GJA
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    Re: Factoring and the Distributive Property Question

    Hi, Srengam. First, kudos for working through this stuff!

    Typically cubics (when the highest power of x is 3) are not easy to factor. However, the one you have can be done if you group the first two terms together x^{3} and 5x^{2}, and group the second two terms -x and -5 together, then factoring these groupings individually.

    If that's not clear, let me know and I will write up what I mean more explicitly. Good luck!
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  4. #4
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    Re: Factoring and the Distributive Property Question

    Quote Originally Posted by Srengam View Post
    But I've come across a question half way through that I can't solve, I'm not sure how to tackle it, the question and answer are
    x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)
    Note that x^{3}+5x^{2}-x-5=x^2(x+5)-(x+5)=(x^2-1)(x+5)
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  5. #5
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    Re: Factoring and the Distributive Property Question

    Thanks all,

    I can see now what I need to "pull out" to simplify it all.

    Thanks again.
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