Answer
$(2x+3)^2$
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 $a=4$, $b=12$, and $c=9$.
Thus, $ac = 4(9) = 36$
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:
$4x^2+12x+9 =4x^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:
$=(4x^2+6x)+(6x+9)
\\=2x(2x+3) +3(2x+3)$
Factor out the GCF $2x+3$ to obtain:
$=(2x+3)(2x+3)
\\=(2x+3)^2$
