Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 326: 114

Answer

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Work Step by Step

Intercepts: $(-7,0)$, $(-.5,0)$, $(4,0)$, $(0,-1)$ $x=-7$ $x+7=-7+7$ $x+7=0$ $x=-.5$ $x+.5=-.5+.5$ $x+.5=0$ $x=4$ $x-4=4-4$ $x-4=0$ $(x+7)(x+.5)(x-4)$ $(x+7)(x-4)(x+.5)$ $(x*x+x*-4+7*x+7*-4)(x+.5)$ $(x^2-4x+7x-28)(x+.5)$ $x^2*x+.5*x^2+(-4x)*x+(-4)*.5+7x*x+7x*.5+(-28)(x)+(-28)(.5)$ $x^3+.5x^2-4x^2-2+7x^2+3.5x-28x-14$ $x^3+(.5x^2-4x^2+7x^2)+(3.5x-28x)-14-2$ $x^3+3.5x^2-24.5x-16$ The resulting function will have some parts of this expression.
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