Answer
$(2z+w)^2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
4z^2+4zw+w^2
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
In the given expression the value of $ac$ is $
4(1)=4
$ and the value of $b$ is $
4
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,4\}, \{2,2\},
\{-1,-4\}, \{-2,-2\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
2,2
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
4z^2+2zw+2zw+w^2
.\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}
(4z^2+2zw)+(2zw+w^2)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2z(2z+w)+w(2z+w)
.\end{array}
Factoring the $GCF=
(2z+w)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2z+w)(2z+w)
\\\\=
(2z+w)^2
.\end{array}