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