Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 3 - Polynomial and Rational Functions - Section 3.3 Complex Zeros; Fundamental Theorem of Algebra - 3.3 Assess Your Understanding - Page 232: 51

Answer

$6x^3-13x^2-13x+20$

Work Step by Step

We simplify the given equation as follows: $2x(3x^2+x-4)-5(3x^2+x-4)\\=2x(3x^2) + 2x(x) - 2x(4) - 5(3x^2)-5(x)-5(-4) \\=6x^3+2x^2-8x-15x^2-5x+20$ Finally, we combine the like terms to obtain: $6x^3+2x^2-8x-15x^2-5x+20=6x^3 + (2x^2-15x^2)+(-8x-5x)+20 \\=6x^3-13x^2-13x+20$
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