# Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.4 - Factoring Trinomials of the Form ax^2+bx+c - Problem Set 6.4 - Page 261: 11

$2(2n-3)(n+8)$

#### Work Step by Step

Since $2$ is common to all the terms of the equation, we take it out as a common factor: $4n^{2}+26n-48$ =$2(2n^{2}+13n-24)$ In accordance with the rules of factoring trinomials, we need to find two integers whose product is $-48$ and whose sum is $13$. A little searching determines that these numbers are $-3$ and $16$. Therefore, we can express the middle term $13n$ as $(-3n+16n)$ and proceed to factoring by grouping: $2(2n^{2}+13n-24)$ =$2(2n^{2}-3n+16n-24)$ =$2[n(2n-3)+8(2n-3)]$ =$2[(2n-3)(n+8)]$ =$2(2n-3)(n+8)$

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