Question 1026338
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<pre>
                                           x
                    -----------------------------
x^3 + 0x^2 + 0x + 5 | x^4 + 0x^3 + 0x^2 + 2x + 3
                      x^4 + 0x^3 + 0x^2 + 5x
                      ---------------------------
                                        - 3x + 3
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Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^4\ +\ 2x\ +\ 3}{x^3\ +\ 5}\ =\ x\ +\ \frac{3\ -\ 3x}{x^3\ +\ 5}]


Since *[tex \LARGE h\cdot l\ =\ k\ \Right\ l\ =\ \frac{k}{h}], we can, from the above, say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l(x)\ =\ x\ +\ \frac{3\ -\ 3x}{x^3\ +\ 5}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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