Answer
$(3k-2)^2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
9k^2-12k+4
,$ 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.
$\bf{\text{Solution Details:}}$
To factor the trinomial expression above, note that the value of $ac$ is $
9(4)=36
$ and the value of $b$ is $
12
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,36 \}, \{ 2,18 \}, \{ 3,12 \}, \{ 4,9 \}, \{ 6,6 \},
\\
\{ -1,-36 \}, \{ -2,-18 \}, \{ -3,-12 \}, \{ -4,-9 \}, \{ -6,-6 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-6,-6
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
9k^2-6k-6k+4
.\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}
(9k^2-6k)-(6k-4)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3k(3k-2)-2(3k-2)
.\end{array}
Factoring the $GCF=
(3k-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3k-2)(3k-2)
\\\\=
(3k-2)^2
.\end{array}
