Answer
$5(y-1)(4y-1)$
Work Step by Step
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}
