Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 63


$f(x)(x-1-4i)(x-1+4i)$ The zeros: $x=1+4i$ $x=1-4i$

Work Step by Step

$x^2-2x+17=0~~$ ($a=1,b=-2,c=17$): $x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-2)±\sqrt {(-2)^2-4(1)(17)}}{2(1)}=\frac{2±\sqrt {-64}}{2}=\frac{2±8i}{2}=1±4i$ $x=1+4i$ or $x=1-4i$ $f(x)=[x-(1+4i)][x-(1-4i)]=(x-1-4i)(x-1+4i)$
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