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Math Help - factoring a higher order trinomial (odd)

  1. #1
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    factoring a higher order trinomial (odd)

    Hi - I am stuck on a question that requires me to find the 2nd point of intersection between a power function (x^3) and the tangent of the function at a given point (1/2, 1/8), which I determined to be y=3/4x - 1/4

    I have combined the two equations and created a 3rd order trinomial 0=x^3 -3/4x +1/4. My problem is that I forget how to factor a trinomial with an odd power. Any help would be appreciated.

    Thanks in advance.
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  2. #2
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    Re: factoring a higher order trinomial (odd)

    Honestly, its really hard to remember the cubic one,

    but if you need it solved, click the link.

    0=x^3 -3/4x +1/4 - Wolfram|Alpha
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  3. #3
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    Re: factoring a higher order trinomial (odd)

    Thanks for the link - I see it being of value in the future. However, I really want to figure this one out.
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  4. #4
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    Re: factoring a higher order trinomial (odd)

    But you know that this line is a tangent to the cubic so you have a repeated root at x=1/2.


    Now you just need to find the other factor.
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  5. #5
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    Re: factoring a higher order trinomial (odd)

    Thanks a tutor. I understand that the first point of intersection is 1/2 and that the tangent will intersect at a second point, which can be determined by figuring out the second factor, but that is exactly my problem. I cannot remember how to determine the other factor for a cubic trinomial that cannot be simplified into a quadratic. I believe mine cannot 0=x^3 -(3/4)x + 1/4.

    Honestly, it has been a while since I studied these earlier concepts and I used to struggle with factoring. Is there a different theorem or method that can be applied?
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  6. #6
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    Re: factoring a higher order trinomial (odd)

    0=x^3 -(3/4)x + 1/4
    x = \frac{1}{2} is a zero, use synthetic division ...

    Code:
    [1/2].........1.........0........-3/4..........1/4
    .......................1/2........1/4..........-1/4
    -------------------------------------------------------
    ..............1........1/2.......-1/2............0
    x^3 - \frac{3}{4}x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x^2 + \frac{1}{2}x - \frac{1}{2}\right)
    Thanks from slowlyGettingIt
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  7. #7
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    Re: factoring a higher order trinomial (odd)

    I was thinking

    x^3-\frac{3}{4}x+\frac{1}{4}=(x-1/2)^2(?+?)

    It's pretty clear that it has to be x^3-\frac{3}{4}x+\frac{1}{4}=(x-1/2)^2(x+1)

    I think I'm lazy.
    Thanks from slowlyGettingIt and HallsofIvy
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  8. #8
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    Re: factoring a higher order trinomial (odd)

    Thanks guys. You both have helped a lot.
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