Question 1188724
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -x^4\ -\ 3x^2\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -\(x^4\ +\ 3x^2\ +\ 2\)]


Let *[tex \Large u\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -\(u^2\ +\ u\ +\ 2\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -(u\ +\ 1)(u\ +\ 2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -(x^2\ +\ 1)(x^2\ +\ 2)]


Let *[tex \Large x^2\ +\ 1\ =\ 0], then the factors of *[tex \Large x^2\ +\ 1] are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{0\ +\ \sqrt{0^2\,-\,(4)(1)(1)}}{2}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{0\ -\ \sqrt{0^2\,-\,(4)(1)(1)}}{2}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ i]


Similarly the factors of *[tex \Large x^2\ +\ 2] are *[tex \Large x\ -\ i\sqrt{2}] and *[tex \Large x\ +\ i\sqrt{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ \in\ \{i,-i,i\sqrt{2},-i\sqrt{2}\}]


Therefore 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ \not\in\ \mathbb{R}]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

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