Answer
$x^3-4x^2-3x+12=0$
Work Step by Step
$x=\sqrt{3}$ and $x=-\sqrt{3}$ and $x=4$
$x-\sqrt{3}=0$ and $x+\sqrt{3}=0$ and $x-4=0$
$(x-\sqrt{3})(x+\sqrt{3})(x-4)=0$
Therefore, multiplying out the factors.
$(x^2-3)(x-4)=0$,
$x^3-4x^2-3x+12=0$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 1 - 1.6 - Other Types of Equations - 1.6 Exercises - Page 129: 87
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