Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 53



Work Step by Step

Given: $-4x[x(3x-2)-8](25x^2-40x+16)=0$ Need to use factorization to solve the quadratic equation. This can be factorized as follows: $-4x(3x+4)(x-2)(5x-4)^2=0$ Now $-4x=0 \implies x=0$ and $(3x+4)=0$ This implies $x=\dfrac{-4}{3}$ and $(x-2)=0 \implies x=2$ and $(5x-4)^2=0$ This implies $x=\dfrac{4}{5}$ Hence, $x=0,\dfrac{-4}{3},2,\dfrac{4}{5}$
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