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Thread: Polynomial equation

  1. #1
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    Polynomial equation

    Determine the polynomial equation in standard form that has the following roots:

    -1 (of order 2), and (2+2sqrt 3) and (2-2sqrt 3)

    please show all work!
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  2. #2
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    Hello, meli3000!

    Determine the polynomial equation in standard form that has the following roots:

    . . $\displaystyle -1\:\text{(of order 2)},\;\,2+2\sqrt{3},\;\;2-2\sqrt{3}$

    If the polynomial $\displaystyle P(x)$ has root $\displaystyle -1\text{ (order 2)}$, then $\displaystyle (x+1)^2$ is a factor of $\displaystyle P(x)$.

    If $\displaystyle 2 + 2\sqrt{3}$ is a root, then $\displaystyle x - (2 + 2\sqrt{3})$ is a factor.

    If $\displaystyle 2 - 2\sqrt{3}$ is a root, then $\displaystyle x - (2 - 2\sqrt{3})$ is a factor.


    Hence: .$\displaystyle P(x) \;=\;(x+1)^2(x - [2+2\sqrt{3}])(x - [2-2\sqrt{3}]) \;=\;0$

    And we have: .$\displaystyle (x^2 + 2x + 1)(x^2-4x - 8) \;=\;0$

    . . Therefore: . $\displaystyle \boxed{x^4 - 2x^3 - 15x^2 - 20x - 8 \;=\;0}$

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