Answer
$(3x-2)^2$
Work Step by Step
RECALL:
A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$
If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping.
The given trinomial has $a=9$, $b=-12$, and $c=4$.
Thus, $ac = 9(4) = 36$
Note that $36=-6(-6)$ and $(-6)+(-6)=-12$
This means that $d=-6$ and $e=-6$.
Rewrite the middle term of the trinomial as $-6x$ +($-6x$) to obtain:
$9x^2-12x+4 =9x^2-6x+(-6x)+9 $
Group the first two terms together and the last two terms together.
Then, factor out the GCF in each group to obtain:
$=(9x^2-6x)+(-6x+4)
\\=3x(3x-2) +(-2)(3x-2)$
Factor out the GCF $3x-2$ to obtain:
$=(3x-2)(3x-2)
\\=(3x-2)^2$
