To find factor of a polynomial equation 
One of the factors of $4x^2+y^2+14x-7y-4xy+12$ is equal to

*

*$2x-y+4$


*$2x-y-3$


*$2x+y-4$


*$2x-y+3$

Step $1$:
$4x^2+y^2-4xy$ can be simplified as $(2x-y)^2$
Step $2$:
$14x-7y$ can be simplified as $7(2x-y)$
and finally
$(2x-y) (2x-y+7) + 12$
I can able to factor to this extent only. however can't able to arrive at the answer.
The answer is given in the book. it states that $4x^2+y^2+14x-7y-4xy+12$ is product of $(2x-y+3)$ and $(2x-y+4)$ I am in need of steps

We can write the given equation as the product of two smaller polynomials:
\begin{align}
  \rlap{4x^2+y^2+14x−7y−4xy+12} \\
  &= (ax+by+c)(dx+ey+f) \\
  &= adx^2+aexy+afx+bdxy+bey^2+bfy+cdx+cey+cf \\
  &= \rlap{adx^2+bey^2+(af+cd)x+(bf+ce)y+(ae+bd)xy+cf}
\end{align}
We can equate coefficients to get the following:
\begin{align}4 &= ad \tag{$x^2$ term} \\
1  &= be \tag{$y^2$ term} \\
14 &= af+cd \tag{$x$ term} \\
-7 &= bf+ce \tag{$y$ term} \\
-4 &= ae+bd \tag{$xy$ term} \\
12 &= cf \tag{constant term}\end{align}
We can solve these simultaneous equations to get 
\begin{align}a &= 2  \\
b &= -1 \\
c &= 4  \\
d &= 2  \\
e &= -1 \\
f &= 3\end{align}
Therefore, we can say that $$4 x^2 + y^2 + 14 x − 7 y − 4 x y + 12 = (2 x - y + 4) (2 x - y + 3)$$
