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

$5(y-1)(4y-1)$
Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 20y^2-25y+5 \\\\= 5(4y^2-5y+1) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(4y^2-5y+1) \end{array} has $ac= 4(1)=4$ and $b= -5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -4,-1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(4y^2-4y-1y+1) .\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} 5[(4y^2-4y)-(y-1)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[4y(y-1)-(y-1)] .\end{array} Factoring the $GCF= (y-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(y-1)(4y-1)] \\\\= 5(y-1)(4y-1) .\end{array}