Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises - Page 134: 64

Answer

The solutions are $x=\sqrt{5},$ $x=-\sqrt{5}$ and $x=2$

Work Step by Step

$x^{3}-2x^{2}-5x+10=0$ On the left side of the equation, group the first two and the last two terms together: $(x^{3}-2x^{2})-(5x-10)=0$ Take out common factor $x^{2}$ from the first parentheses and common factor $5$ from the second one: $x^{2}(x-2)-5(x-2)=0$ Take out common factor $x-2$ $(x-2)(x^{2}-5)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x^{2}-5=0$ $x^{2}=5$ $x=\pm\sqrt{5}$ The solutions are $x=\sqrt{5},$ $x=-\sqrt{5}$ and $x=2$
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