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Math Help - Problems concerning polynomials / complex numbers

  1. #1
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    Problems concerning polynomials / complex numbers

    f(x) = |x+5| + |x+2| + |x| + |x-3|

    The equation gives us the total length of a cable between the points A and B

    P=-5
    Q= -2
    O=0
    R= 3

    x is the centre of the cable.

    The question is, where should the centre of the cable be to minimize toe total length of cable to all four points. And: what is the minimum (of the length of the cable)

    Also, how can I graph this function that has absolute values in its equation?

    Last part of this problem: A point S is added 7km away from O (so it adds to the equation + |x-7|. Where should it be for a minimal cable length?




    Next question:
    I need to find a fourth degree polynomial with the rational coefficients 2- i*squareroot 3 and squareroot2 +1 as two of the four zeros. How do I do this and what is the solution? Please help me.
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  2. #2
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    Hello, Instigator!

    Here's #2 . . .


    2) Find a fourth-degree polynomial with the rational coefficients
    . . where 2 - i\sqrt{3} and 1 + \sqrt{2} are two of the four zeros.

    You're expected to know these facts . . .

    If 2 - i\sqrt{3} is a zero, then its conjugate, 2 + i\sqrt{3}, is also a zero.

    If 1 + \sqrt{2} is a zero, then its conjugate, 1 - \sqrt{2}, is also a zero.


    The polynomial is: . P(x) \;=\;\left[x - (2-i\sqrt{3})\right]\cdot\left[x - (2+i\sqrt{3})\right]\cdot\left[x-(1 + \sqrt{2})\right]\cdot\left[x - (1-\sqrt{2})\right]


    Multiply and simplify: . \boxed{P(x) \;\;=\;\;x^4 - 6x^3 + 14x^2 + 10x - 7}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    If the thought of that multiplication fills you with Fear and Loathing,
    . . let me show you a trick for handling "conjugate multiplication".


    The first product is: . \left[x - (2 - i\sqrt{3})\right]\cdot \left[x - (2 + i\sqrt{3})\right]

    . . Regroup it like this: . \left[(x-2) + i\sqrt{3}\right]\cdot\left[(x-2) - i\sqrt{3}\right] . . . This is: (a+b)(a-b)

    . . So we have: . (x-2)^2 - (i\sqrt{3})^2 \;\;=\;\;x^2 - 4x + 4 -(-3) \;\;=\;\;{\color{blue}x^2-4x + 7}


    The second product is: . \left[x - (1 + \sqrt{2})\right]\cdot\left[x - (1 - \sqrt{2})\right]

    . . Regroup it like this: . \left[(x-1) - \sqrt{2}\right]\cdot\left[(x-1) + \sqrt{2}\right] . . . This is: a^2-b^2

    . . So we have: . (x-1)^2 - (\sqrt{2})^2 \;\;=\;\;x^2-2x+1-2 \;\;=\;\;{\color{blue}x^2-2x-1}


    Then we can multiply: . \bf{{\color{blue}(x^2-4x+7)(x^2-2x-1)}} . . . see?

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  3. #3
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    Thank you very much, especially for the trick and the great explanation!
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