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Math Help - Polynomial Functions help please!!!

  1. #1
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    Smile Polynomial Functions help please!!!

    How would i write a polynomial equation from the roots 3 and 2+√7?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by peepsrock09 View Post
    How would i write a polynomial equation from the roots 3 and 2+√7?
    Note that x=3 is a zero, which implies x-3 is a factor of the polynomial

    Note that x=2+\sqrt{7} is a zero, which implies x-\left(2+\sqrt{7}\right) is a factor of the polynomial. We also need the conjugate of our previous zero x=2-\sqrt{7} to keep the coefficients clear of square roots. That implies that x-\left(2-\sqrt{7}\right) is a factor of the polynomial.

    Thus, your polynomial as a product of factors is \left(x-3\right)\left(x-\left[2+\sqrt{7}\right]\right)\left(x-\left[2-\sqrt{7}\right]\right)

    Foil this out and see what you get.

    Does this make sense?
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  3. #3
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    I came up with x^3-8x^2+16x-3....

    this is correct? or did i foil wrong?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by peepsrock09 View Post
    I came up with x^3-8x^2+16x-3....

    this is correct? or did i foil wrong?
    Unfortunately, this is not correct.

    How did you foil it out?? If you show what you did, I can easily show you where you made your mistake.
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  5. #5
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    i took that equation, foiled to get:
    (x-3)[x^2-x(2-√ 7)-x(2+√ 7)+(2+√ 7)(2-√ 7)]
    (x-3)(x^2-2x+√ 7x-x-2x+1-√ 7x)
    (x-3)(x^2-5x+1)
    =
    x^3-8x^2+16x-3



    I think i foiled wrong in the first place..

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  6. #6
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    Quote Originally Posted by peepsrock09 View Post
    i took that equation, foiled to get:
    (x-3)[x^2-x(2-√ 7)-x(2+√ 7)+(2+√ 7)(2-√ 7)]
    (x-3)(x^2-2x+√ 7x-x-2x+1-√ 7x)
    (x-3)(x^2-5x+1)
    =
    x^3-8x^2+16x-3



    I think i foiled wrong in the first place..

    A simpler approach:

    \left( x - \left[ 2 + \sqrt{7} \right] \right) \left(x - \left[ 2 - \sqrt{7} \right] \right) = \left( [x - 2] - \sqrt{7} \right) \left( [x-2] + \sqrt{7} \right)

    which has the form of the factorisation of the difference of two squares:

    (A - B)(A + B) = A^2 - B^2

    (and yes I am attempting the record of how many times I can use the word of in one sentence).

    Identifying A = (x - 2) and B = \sqrt{7} you therefore have:

    (x - 2)^2 - \left( \sqrt{7} \right)^2 = x^2 - 4x + 4 - 7 = x^2 - 4x - 3.
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  7. #7
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    aaahhhh. thank you soo much
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