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Math Help - Factor 3x^2 +2x -6

  1. #1
    Member nautica17's Avatar
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    Factor 3x^2 +2x -6

    I'm kind of embarrassed to ask this question since I am in calculus but how would I factor the following? I know you can multiply 3*-6 and you get -18 and somehow you then can factor. But I can't seem to do it.

    3x^2 + 2x -6

    Any help please?
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  2. #2
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    Quote Originally Posted by nautica17 View Post
    I'm kind of embarrassed to ask this question since I am in calculus but how would I factor the following? I know you can multiply 3*-6 and you get -18 and somehow you then can factor. But I can't seem to do it.

    3x^2 + 2x -6

    Any help please?
    b^2 - 4ac = 4 - 4(3)(-6) = 4 + 72 = 76

    since the discriminant is not a perfect square number, the quadratic will not factor.
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  3. #3
    Member nautica17's Avatar
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    Quote Originally Posted by skeeter View Post
    b^2 - 4ac = 4 - 4(3)(-6) = 4 + 72 = 76

    since the discriminant is not a perfect square number, the quadratic will not factor.
    Hmm... I see. Well, I'm trying to find the local extrema and that equation happens to be the second derivative. I don't know how to get the max and min point out of that now.
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    Quote Originally Posted by nautica17 View Post
    Hmm... I see. Well, I'm trying to find the local extrema and that equation happens to be the second derivative. I don't know how to get the max and min point out of that now.
    you don't need to know where f''(x) = 0 to find extrema ... inflection points, yes, but not extrema.

    if f'(x) = 0 and f''(x) > 0 , you have a min

    if f'(x) = 0 and f''(x) < 0 , you have a max
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  5. #5
    Member nautica17's Avatar
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    Okay, so my f'(x) = x^3 + x^2 -6
    From that I got x(x+3)(x-2)

    So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?
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  6. #6
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    Quote Originally Posted by nautica17 View Post
    Okay, so my f'(x) = x^3 + x^2 -6
    From that I got x(x+3)(x-2)

    So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?
    check f'(x) again ... did you mean to type f'(x) = x^3 + x^2 - 6x ?

    it might be a good idea to see the original function, f(x).
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  7. #7
    Member nautica17's Avatar
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    Quote Originally Posted by skeeter View Post
    check f'(x) again ... did you mean to type f'(x) = x^3 + x^2 - 6x ?

    it might be a good idea to see the original function, f(x).

    Yea, that's what I meant, sorry.

    The original function was ((x^4)/4)+((x^3)/3)-(3x^2)+5
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  8. #8
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    Quote Originally Posted by nautica17 View Post
    Okay, so my f'(x) = x^3 + x^2 -6x
    From that I got x(x+3)(x-2)

    So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?
    sorry, no.

    min at x = -3

    max at x = 0

    min at x = 2
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  9. #9
    Member nautica17's Avatar
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    Hmm, okay cool, thank-you. I think I get it now.
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