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

$-(2x^{n}+1)(10x^n+3)$
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}