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: 64


$f(x)=(x+5-2\sqrt 2)(x+5+2\sqrt 2)$ The zeros: $x=-5+2\sqrt 2$ $x=-5-2\sqrt 2$

Work Step by Step

$x^2+10x+17=0~~$ ($a=1,b=10,c=17$): $x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-10±\sqrt {10^2-4(1)(17)}}{2(1)}=\frac{-10±\sqrt {32}}{2}=\frac{-10±4\sqrt 2}{2}=-5±2\sqrt 2$ $x=-5+2\sqrt 2$ or $x=-5-2\sqrt 2$ $f(x)=[x-(-5+2\sqrt 2)][x-(-5-2\sqrt 2)]=(x+5-2\sqrt 2)(x+5+2\sqrt 2)$
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