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Math Help - Having trouble Factoring

  1. #1
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    Having trouble Factoring

    Here's my question.

    Determine the x-intercepts of the polynomial function.

    6x^3 + 13x^2 +2x

    So I've factored it down to

    x(6x^2 + 13x + 2)

    Do I have to find 2 numbers that are the product of 2 and the sum of 13? I don't think it's possible. What do I do?
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by Arise01 View Post
    Here's my question.

    Determine the x-intercepts of the polynomial function.

    6x^3 + 13x^2 +2x

    So I've factored it down to

    x(6x^2 + 13x + 2)

    Do I have to find 2 numbers that are the product of 2 and the sum of 13? I don't think it's possible. What do I do?
    Good start.

    To factor this won't be easy. To start with, you have a 6x^2 to consider.

    This means you could have something of the form (3x+...)(2x+...) or you could have something of the form (6x+...)(x+...) as either will generate a 6x^2 when expanded.

    The way to do this is to write out the factors of 6:

    2,3
    1,6

    And write out the factors of 2:

    1,2

    And find combinations that make 13 by multiplying pairs of numbers and adding them. It's basically trial and error here.

    Starting with '2,3' and '1,2':

    2 x 1 + 3 x 2 = 8
    2 x 2 + 3 x 1 = 7

    Neither combination is going to work. Note that 2 x 3 + 1 x 2 is not a combination to consider.

    So it must be from 1,6 and 1,2
    1 x 6 + 1 x 2 = 8
    1 x 1 + 6 x 2 = 13

    So there's the combination we need. Then you need to think about the correct arrangements. You need to multiply 6 and 2, so they need to be in opposing brackets. This means the only possible result is:
    (x+2)(6x+1)
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  3. #3
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    So it would be x(x+2)(6x+1) ?
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  4. #4
    Super Member Quacky's Avatar
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    Quote Originally Posted by Arise01 View Post
    So it would be x(x+2)(6x+1) ?
    Remember that it's an equation you need, not an expression. You must write x(x+2)(6x+1)=0.
    ...So what are your three solutions this time?
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  5. #5
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    x=0, x+2=0, 6x+1=0

    So 0, -2, 1/6 are the x-intercepts?
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  6. #6
    Super Member Quacky's Avatar
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    Quote Originally Posted by Arise01 View Post
    x=0, x+2=0, 6x+1=0

    So 0, -2, 1/6 are the x-intercepts?
    Nearly, there's a sign error:

    If 6x+1=0
    then 6x=-1
    So x=-1/6

    The other two are fine.
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  7. #7
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    Thanks again quacky.
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