Question 222298
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Start with the rational root theorem.  The possible rational roots are all *[tex \Large \frac{p}{q}] where *[tex \Large p] is an integer factor of the constant term and *[tex \Large q] is an integer factor of the high-order term.  That means that the possible rational roots for:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3 - 4x^2 - 2x + 20\ =\ 0]


are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm1, \pm2, \pm4, \pm5, \pm10, \pm20]


Next you need to use synthetic division to determine which of the 12 possible rational zeros are actually zeros.  If you need a tutorial on synthetic division, Purplemath has a very good one at http://www.purplemath.com/modules/synthdiv.htm


I'm going to take a shortcut here and tell you that -2 is a zero, hence *[tex \Large x + 2] is a factor of *[tex \Large x^3 - 4x^2 - 2x + 20] and dividing *[tex \Large x^3 - 4x^2 - 2x + 20] by *[tex \Large x + 2] leaves a quotient of *[tex \Large x^2 - 6x + 10]


So now we have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3 - 4x^2 - 2x + 20\ =\ (x + 2)(x^2 - 6x + 10)]


Leaving us with having to solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + 10\ =\ 0]


which does not factor so we need to use the quadratic formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-6) \pm sqrt{(-6)^2 - 4(1)(10)}}{2(1)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{6 \pm sqrt{-4}}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{6 \pm 2i}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 3 \pm 2i ]


Therefore the three roots of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3 - 4x^2 - 2x + 20\ =\ 0]


are *[tex \LARGE -2], *[tex \LARGE 3 + 2i], and *[tex \LARGE 3 - 2i]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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