How do I factorize this? $8 + 15xy − 12x − 10y $
How would I go about factorizing this? I'm not sure if it is even possible.
 A: If the expression factors at all, its factorization must be of the form $15(x-b)(y-c)$.
Setting $y=0$ in the original equation leaves $8-12x=0$ so the root in $x$ must be $b=\frac{2}{3}\,$. Similarly, setting $x=0$ gives $c=\frac{4}{5}\,$.
Then, the tentative factorization is $15\left(x-\frac{2}{3}\right)\left(y-\frac{4}{5}\right) = (3x-2)(5y-4)\,$.
Since this was derived on the assumption that such a factorization does in fact exist, the result must be verified, and it is indeed easily verified that this is the correct factorization.
A: $$8 + 15xy − 12x − 10y$$
$$=15xy-10y+8-12x$$
$$=5y(3x-2)+8-12x$$
$$=5y(3x-2)+4(2-3x)$$
$$=5y(3x-2)-4(3x-2)$$
$$=(5y-4)(3x-2)$$
A: Observe that the $15xy$ and $-10y$ terms both have $5y$ as a common factor, so
$$ 8 + 15xy - 12x - 10y = (3x-2) 5y + 8 - 12 x
$$
Observe that also the $8$ and $-12x$ terms have $-4$ as a common factor, so
$$ (3x-2) 5y + 8 - 12 x = (3x-2)5y - (3x-2)4
$$
Then
$$ (3x-2)5y - (3x-2)4 = (3x-2)(5y-4)
$$
A: We can write $$P= (15xy-12x) +(8-10y) =(3x)(5y-4) +(-2)(5y-4) =(3x-2)(5y-4) $$ Hope it helps. 
A: Though you have already got many answers, but alternately it can be done multiplying and dividing the entire expression by $2$ to simplify it, i.e.,
$8+15xy-12x-10y = [15xy-10y-12x+8]\times2/2$
$=[\frac{15xy}{2}-5y-6x+4]\times2$
$=[5y(\frac{3x}{2}-1)-2(3x-2)]\times2$
$=[5y(3x-2)-4(3x-2)]$
$=(5y-4)(3x-2)$
A: There is also the possibility of going brute force on it.
$(ax+b)(cy+d)=acxy+adx+bcy+bd$.
Identifying the coefficients to $8+15xy−12x−10y\;$ we get 
$\begin{cases}
bd=8\\
ac=15\\
ad=-12\\
bc=-10\\
\end{cases}$
Notice that $(aqx+bq)(\frac{c}{q}y+\frac{d}{q})=(ax+b)(cy+d)$ so we can fix any of $a,b,c,d$ to a desired value by multiplying by a convenient rational $q$.
Let's have for instance $a=3$ which seems attractive.
The system immediately solves to $\begin{cases}
a=3\\
b=-2\\
c=5\\
d=-4\\
\end{cases}$
And $(3x-2)(5y-4)$ is the desired factorization.
