Question 863068
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Not all of the remaining zeros are complex.  Do synthetic division on the original 5th degree equation polynomial using -3 as a divisor.  The result is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)(x^4\ -\ 2x^3\ +\ 5x^2\ -\ 8x\ +\ 4)\ =\ 0]


The possible real roots of the quartic are *[tex \LARGE \pm1,\ \pm2,\ \pm4].


Use synthetic division to find two more real roots.  Hint:  The two real roots are equal.


You will be left with a quadratic that you can solve using the quadratic formula for the conjugate pair of complex roots.


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