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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 75

#### Answer

$x^2(2x+3)(7x-1)$

#### Work Step by Step

Factoring the $GCF= x^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 19x^3-3x^2+14x^4 \\\\= 14x^4+19x^3-3x^2 \\\\= x^2(14x^2+19x-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} x^2(14x^2+19x-3) \end{array} has $ac= 14(-3)=-42$ and $b= 19 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 21,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2(14x^2+21x-2x-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} x^2[(14x^2+21x)-(2x+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2[7x(2x+3)-(2x+3)] .\end{array} Factoring the $GCF= (2x+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} x^2[(2x+3)(7x-1)] \\\\= x^2(2x+3)(7x-1) .\end{array}

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