Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - Review Exercises - Page 154: 83


Solutions: $x=0$ $x=-\frac{2}{3}$ $x=+\frac{2}{3}$ $x=-3$

Work Step by Step

$9x^4+27x^3-4x^2-12x=0$ $9x^3(x+3)-4x(x+3)=0$ $(9x^3-4x)(x+3)=0$ $x(9x^2-4)(x+3)=0$ $x[(3x)^2-2^2](x+3)=0$ $x(3x+2)(3x-2)(x+3)=0$ Solutions: $x=0$ $3x+2=0$ $x=-\frac{2}{3}$ $3x-2=0$ $x=+\frac{2}{3}$ $x+3=0$ $x=-3$ Check the solutions: $x=0$: $9(0)^4+27(0)^3-4(0)^2-12(0)=0$ $x=-\frac{2}{3}$ $9(-\frac{2}{3})^4+27(-\frac{2}{3})^3-4(-\frac{2}{3})^2-12(-\frac{2}{3})=9\frac{16}{81}-27\frac{8}{27}-4\frac{4}{9}+8=0$ $x=\frac{2}{3}$ $9(\frac{2}{3})^4+27(\frac{2}{3})^3-4(\frac{2}{3})^2-12(\frac{2}{3})=9\frac{16}{81}+27\frac{8}{27}-4\frac{4}{9}-8=0$ $x=-3$: $9(-3)^4+27(-3)^3-4(-3)^2-12(-3)=9(81)-27(27)-4(9)+36=0$
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