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 293: 17

Answer

(a) $\pm1,\frac{1\pm\sqrt 3i}{2},\frac{-1\pm\sqrt 3i}{2}$ (b) $P(x)=(x+1)(x-1)(x-\frac{1+\sqrt 3i}{2})(x-\frac{1-\sqrt 3i}{2})(x+\frac{1-\sqrt 3i}{2})(x+\frac{1+\sqrt 3i}{2})$

Work Step by Step

$P(x)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ (a) Find the zeros for the quadratics, we have all the zeros as $\pm1,\frac{1\pm\sqrt 3i}{2},\frac{-1\pm\sqrt 3i}{2}$ (b) $P(x)=(x+1)(x-1)(x-\frac{1+\sqrt 3i}{2})(x-\frac{1-\sqrt 3i}{2})(x+\frac{1-\sqrt 3i}{2})(x+\frac{1+\sqrt 3i}{2})$
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