Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.2 Factoring Trinomials of the Type x2+bx+c - 5.2 Exercise Set: 43

Answer

$3x(x-25)(x+4)$

Work Step by Step

Factoring the $GCF= 3x $, the given expression, $ 3x^3-63x^2-300x ,$ is equivalent to \begin{array}{l} 3x(x^2-21x-100) .\end{array} The 2 numbers whose product is $ac= 1(-100)=-100 $ and whose sum is $b= -21 $ are $\left\{ -25,4 \right\}.$ Using these two numbers to decompose the middle term of the above trinomial, then \begin{array}{l} 3x(x^2-25x+4x-100) \\\\= 3x[(x^2-25x)+(4x-100)] \\\\= 3x[x(x-25)+4(x-25)] \\\\= 3x[(x-25)(x+4)] \\\\= 3x(x-25)(x+4) .\end{array}
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