Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 294: 49

Answer

$1,\dfrac{1 + i\sqrt 3}{2},\dfrac{1 -i\sqrt 3}{2}$

Work Step by Step

Consider $P(x)=x^3-2x^2+2x-1=0$ In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows: $x^3-2x^2+2x-1=(x-1)(x^2-x+1)=0$ This gives: $(x-1)(\dfrac{-(-1)\pm\sqrt{1-4}}{2})=0$ This can be re-written as: $(x-1)(\dfrac{1\pm\sqrt{-3}}{2})=0$ $(x-1)(\dfrac{1\pm i\sqrt 3}{2})=0$ Thus the zeroes of $P(x)$ are: $1,\dfrac{1 + i\sqrt 3}{2},\dfrac{1 -i\sqrt 3}{2}$
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