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 - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 327: 96



Work Step by Step

Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -20x^{2n}-16x^n-3 \\\\= -1(20x^{2n}+16x^n+3) \\\\= -(20x^{2n}+16x^n+3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -(20x^{2n}+16x^n+3) \end{array} has $ac= 20(3)=60 $ and $b= 16 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 10,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -(20x^{2n}+10x^n+6x^n+3) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -[(20x^{2n}+10x^n)+(6x^n+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -[10x^n(2x^{n}+1)+3(2x^n+1)] .\end{array} Factoring the $GCF= (2x^{n}+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} -[(2x^{n}+1)(10x^n+3)] \\\\= -(2x^{n}+1)(10x^n+3) .\end{array}
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