Answer
$2(x+1)(3x+1)$
Work Step by Step
Factor out $2$ to obtain:
$=2(3x^2+4x+1)$
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 trinomial above has $a=3$, $b=4$, and $c=1$.
Thus, $ac = 3(1) = 3$
Note that $3=3(1)$ and $3+1=4$
This means that $d=3$ and $e=1$.
Rewrite the middle term of the trinomial as $3x$ +($x$) to obtain:
$3x^2+4x+1 =3x^2+3x+x+1 $
Group the first two terms together and the last two terms together.
Then, factor out the GCF in each group to obtain:
$=(3x^2+3x)+(x+1)
\\=3x(x+1) +(1)(x+1)$
Factor out the GCF $x+1$ to obtain:
$=(x+1)(3x+1)$
Therefore the completely factored form of the given expression is:
$=2(x+1)(3x+1)$
