## 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: 63

#### Answer

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

#### Work Step by Step

Factoring the negative $GCF= -t^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -14t^4+19t^3+3t^2 \\\\= -t^2(14t^2-19t-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -t^2(14t^2-19t-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} -t^2(14t^2-21t+2t-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} -t^2[(14t^2-21t)+(2t-3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -t^2[7t(2t-3)+(2t-3)] .\end{array} Factoring the $GCF= (2t-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} -t^2[(2t-3)(7t+1)] \\\\= -t^2(2t-3)(7t+1) .\end{array}

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