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Math Help - Problem with factoring.

  1. #1
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    Problem with factoring.

    First off, I do not believe this is a binomial or trinomial; does anyone know what to call this equation?

    Alright, my problem is that I do not understand how this problem goes from one step to the next:

    Step 1) X 2x + x + 5(x-1) = 0
    Step 2) X(x+2x+1) + 5(x-1) = 0
    Step 3) X(x-1) + 5(x-1) = 0
    Step 4) (x+5) (x-1) = 0
    I understand how to get from step one to step two, but what in the world happens to the 2x from step two to step three and how the heck does that +1 turn into a -1?

    Then, in step four where does that x+5 come from? It seems to me that it would be (x+1)(x-1)=0 by the time the equation reaches step four.

    Thanks for any help!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Problem with factoring.

    There's a mistake in step 2, it has to be x^2-2x+1 because you can write:
    x^2-2x+1=(x-1)\cdot (x-1)=(x-1)^2 and this is what happens in step 3.

    In step 3 you have:
    x(x-1)^2+5(x-1)^2, so you have a common factor (x-1)^2 for both terms so you can write:
    x(x-1)^2+5(x-1)^2=(x-1)^2 \cdot (x+5)
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Problem with factoring.

    Quote Originally Posted by Stoneface View Post
    First off, I do not believe this is a binomial or trinomial; does anyone know what to call this equation?

    Alright, my problem is that I do not understand how this problem goes from one step to the next:

    Step 1) X 2x + x + 5(x-1) = 0
    Step 2) X(x+2x+1) + 5(x-1) = 0
    Step 3) X(x-1) + 5(x-1) = 0
    Step 4) (x+5) (x-1) = 0
    I understand how to get from step one to step two, but what in the world happens to the 2x from step two to step three and how the heck does that +1 turn into a -1?

    Then, in step four where does that x+5 come from? It seems to me that it would be (x+1)(x-1)=0 by the time the equation reaches step four.

    Thanks for any help!

    First off, there is a difference between x and X.

    Second off, step 2 is wrong!

    It's should be:

    x(x^2-2x+1)+5(x-1)^2=0

    Third off, remember,

    x^2-2x+1=(x-1)(x-1)=(x-1)^2
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  4. #4
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    Re: Problem with factoring.

    Too many terms (4 if you solve), so it is a polynomial.
    Now to solve from step 1 to 2 "x" is pulled as gcf from first three terms, but there is typo in 2nd step, -2x is changed to +2x which is wrong. But step 3 is correct again where the square is completed as you know (x - 2x + 1) = (x-1)

    Then (x-1) is common and pulled out giving (x + 5) as the other factor.
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  5. #5
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    Re: Problem with factoring.

    Quote Originally Posted by Stoneface View Post
    Step 1) X 2x + x + 5(x-1) = 0
    Step 2) X(x+2x+1) + 5(x-1) = 0
    Step 3) X(x-1) + 5(x-1) = 0
    Step 4) (x+5) (x-1) = 0
    I understand how to get from step one to step two, but what in the world happens to the 2x from step two to step three and how the heck does that +1 turn into a -1?

    Then, in step four where does that x+5 come from? It seems to me that it would be (x+1)(x-1)=0 by the time the equation reaches step four.
    This is quite basic stuff; are you learning on your own?
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