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I have problems following a solution towards simplifying a given polynomial.

$$Polynomial: p(x)=x^5+{\sqrt 3}x^4+24{\sqrt 3}x^2+72x$$

the zeros of this function (Polynomial roots? English isn't my native language, so I don't know how to express the point(s) at which the function meets the X-axis) are: $$x_0 = -{\sqrt 3} \\ x_1 = 0 \\$$ and the complex ones, which are calculated with what "remains" after polynomial division, drawing the 3rd root, etc.: $$x=\sqrt[\Large 3]{-24\sqrt 3}$$

The first problem comes now. The next step, without explanation, simplifies the above to the following: $$x=\sqrt[\Large 3]{8*\sqrt 3^3e^{\Large_{i\pi}}}$$ How is this done or rather what's the logic behind it? Especially the 8 that somehow was transformed from the 24.

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    $\begingroup$ $24 = 8\cdot 3 = 8\cdot(\sqrt{3})^2$ $\endgroup$
    – rogerl
    Commented Feb 16, 2019 at 1:08

2 Answers 2

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$24 = 3*8$ so $24 \sqrt {3} = 8*3*\sqrt{3} = 8\sqrt{3}^3$

$-1 = e^{\pi i}$ so $-24\sqrt{3} = 8*(\sqrt 3)^3 e^{\pi i}=2^3(\sqrt 3)^3e^{\pi i}$

And so $\sqrt[3]{-24\sqrt 3}=\sqrt[3]{2^3\sqrt{3}^3e^{\pi i}} = 2\sqrt 3 e^{\frac {(2k + 1)}3\pi i}$

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  • $\begingroup$ damn so simple. thanks alot. one more small question: how do I calculatethe following zeroes then? I have seen the "formula" for it, which you presented in the end, but I cannot understand it completely, as in some sites there is a φ in the formula, that is somehow calculated (arccos or sth? How am I supposed to calculate that without a calculator? Is it just the division of real and imaginary?) $\endgroup$ Commented Feb 16, 2019 at 1:30
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The $8$ wasn't "transformed from" the $24$. It was factored: $24=8\cdot3=8\cdot\sqrt 3^2$, so $24\sqrt3=8\cdot\sqrt3^3$. The $-1$ became rewritten as $e^{\pi i}$.

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  • $\begingroup$ Thanks alot. One more small question: how do I calculatethe following zeroes then? I have seen the "formula" for it, but I cannot figure out how to give a constantly correct answer to the angle phi (or in some theta). My book shows that it is the absolute value of the complex number z = a+bi, so sqrt(a^2+b^2) but that would give me sqrt(7) in this case, wouldn't it? How do I derive a pi from this? I thought about at which quadrant it would end up in, and said it's either 0 if a & b are positive, pi if a is negative & b positive, -pi if a & b are negative and 0 again if both are negative?? $\endgroup$ Commented Feb 16, 2019 at 2:05

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