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Math Help - Finding a Polynomial Function

  1. #1
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    Finding a Polynomial Function

    Suppose you're given the following zeroes of a polynomial of degree 5, and leading coefficient 1:

    x=2+\sqrt{3}, x=1-i, x=1. Thus, you have four factors as follows (by the complex conjugate theorem): (x-1)(x-(2+\sqrt{3}))(x-(1-i))(x-(1+i)) which more elagantly written is (x-1)((x-2)-\sqrt{3})((x-1)+i)((x-1)-i)..

    Knowing that the polynomial is degree 5 with a leading coefficient of 1, how do we determine the remaiing zero?

    I have some ideas, but I'm not exactly sure. Do we expand out the factors we have and simply make up a factor that will give you a leading co-efficient of 1 and then expand to create the 5th degree polynomial in expanded form? Do we say that there is an additional zero of 2-\sqrt{3}, )?

    Hmm...
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  2. #2
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    Hello, TaylorM0192!

    I think you miscounted the zeroes . . .


    Given these zeroes of a polynomial of degree 5, and leading coefficient 1:

    . . \begin{Bmatrix}2+\sqrt{3} \\ 1-i \\ 1\end{Bmatrix}

    Find the polynomial.

    We already know all five zeros of the polynomial: . \begin{Bmatrix}2 + \sqrt{3} \\ 2-\sqrt{3} \\ 1 - i \\ 1 + i \\ 1 \end{Bmatrix}


    Hence, the polynomial is: . \bigg[x -(2 + \sqrt{3})\bigg]\,\bigg[x - (2 - \sqrt{3})\bigg]\,\bigg[x - (1-i)\bigg]\,\bigg[x - (1+i)\bigg]\,\bigg[x - 1\bigg]

    Just multiply it out . . .


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


    Here's a tip for the multiplication. . .

    Consider: . \bigg[x - (1-i)\bigg]\,\bigg[x-(1+i)\bigg]

    . . . . . . =\;\bigg[(x-1) + i\bigg]\,\bigg[(x-1) - i\bigg]

    . . . . . . =\qquad(x-1)^2 - (i^2)

    . . . . . . =\quad\;\; x^2 - 2x + 1 + 1

    . . . . . . =\qquad\; x^2 - 2x + 2

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