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Math Help - Factoring question using ac method

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    Factoring question using ac method

    3x^2 -4x +1

    The way I understand this is that you multiply 3 and 1 to get 3. I also understand that 3(1) or 1(3) would work by only if they added up to 4 and were able to be multiplied together to make 4 (bx). -3 -1 makes -4. But -3(-1) makes 3. With 3 being the constant I am really struggling to make this thing work.

    The book gives the answer as (3x-1)(x-1)


    Also: I have a similar problem with 2x^2 +11x -9: -2(9) and 2(-9) don't seem to be working either.

    Could be I'm just fried because I'm tired but I'm stumped.
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  2. #2
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    Quote Originally Posted by Ingersoll View Post
    3x^2 -4x +1

    The way I understand this is that you multiply 3 and 1 to get 3. I also understand that 3(1) or 1(3) would work by only if they added up to 4 and were able to be multiplied together to make 4 (bx). -3 -1 makes -4. But -3(-1) makes 3. With 3 being the constant I am really struggling to make this thing work.

    The book gives the answer as (3x-1)(x-1)


    Also: I have a similar problem with 2x^2 +11x -9: -2(9) and 2(-9) don't seem to be working either.

    Could be I'm just fried because I'm tired but I'm stumped.
    We assume there is a solution in integers.

    Based on a = 3, we can reduce to one of these forms

    (3x + u) (x + v)
    (3x + u) (x - v)
    (3x - u) (x + v)
    (3x - u) (x - v)

    for some positive integers u,v.

    Based on c = 1, it's one of these forms

    (3x + u) (x + v)
    (3x - u) (x - v)

    Based on negative b, it's this form

    (3x - u) (x - v)

    The only possibility for u and v are (1,1).

    Next question: 2x^2 +11x -9

    (2x + u) (x + v)
    (2x + u) (x - v)
    (2x - u) (x + v)
    (2x - u) (x - v)

    then

    (2x + u) (x - v)
    (2x - u) (x + v)

    u,v = (1,9) or (3,3) or (9,1).

    You can't get b=11 no matter what you try; this does not factor nicely in integers. (You can still factor using quadratic equation to find the two irrational roots.)
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