Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 5 Polynomials and Polynomial Functions - 5.5 Apply the Remainder and Factor Theorems - 5.5 Exercises - Problem Solving - Page 368: 46


See below.

Work Step by Step

$x=2$ is a solution thus by the factor theorem $x-2$ is a factor of the equation, hence: $-6x^3+72x=96\\-x^3+12x=16\\-x^3+12x-16=0\\x^3-12x+16=0\\(x-2)(x^2+2x-8)=0\\(x-2)(x-2)(x+4)=0$ Thus the solutions are $x=2$ and $x=-4$ but $x\ne2$ and $x\gt0$, thus we have shown that there is no other option.
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