Answer
$(3y+z)^2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
9y^2+6yz+z^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 $
9(1)=9
$ and the value of $b$ is $
6
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,9\}, \{3,3\},
\{-1,-9\}, \{-3,-3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
3,3
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
9y^2+3yz+3yz+z^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}
(9y^2+3yz)+(3yz+z^2)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3y(3y+z)+z(3y+z)
.\end{array}
Factoring the $GCF=
(3y+z)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3y+z)(3y+z)
\\\\=
(3y+z)^2
.\end{array}
