Foiled once again

Factoring give us $xy+3y-4x-12=y(x+3)-4(x+3)=(x+3)(y-4)$ so the sum is $|1+0+3|+|0+1-4|=7$

Given,

$f(x, y) = xy + 3y - 4x - 12$

$f(x, y) = y(x + 3) - 4(x + 3)$

$f(x, y) = (y - 4)(x + 3)$ ...... (1)

Also, it is given that $f(x, y)$ can also be expressed as

$f(x, y) = (ax + by + c)(dx + ey + f)$ ...... (2)

Comparing equations (1) and (2), we have

$a = 0$

$b = 1$

$c = -4$

$d = 1$

$e = 0$

$f = 3$

So, $|a + b + c| + |d + e + f|$

$= |0 + 1 + (-4)| + |1 + 0 + 3|$

$= |-3| + |4|$

$= 3 + 4$

$= 7$

That's the answer!

By Simon's Favorite Factoring Trick, $xy+3y-4x-12=(x+3)(y-4)$ . Therefore our answer is $|1+0+3|+|1+0+-4|=\boxed{7}$

Doing the product we have $f(x,y)=adx^2 + bey^2 + (ae+bd)xy + (af+cd)x + (bf +ce)y + cf=xy + 3y-4x-12$ . Thus $af+cd=-4 \\ bf+ce=3\\ ae+bd=1\\ cf=-12$ the last equation has four cases $c=\pm3, f=\mp4$ but doesn't matter because we want the absolute value so we are going to say that $c=3, f=-4$ and the system of equations is going to be like this, $-4a+3d=-4 \\ -4b+3e=3 \\ ae+bd=1,$ but $ad=be=0,$ because we don't have square terms and say that $a=e=0.$ Substituting $3d=-4 \\ -4b=3\\ bd=1,$ from the last equation we solved it and obtein that $b=d=\pm1,$ because the problem says that the coefficients are integers. So $a=0 \\ b= \pm1 \\ c=\pm3 \\ d=\pm1 \\ e=0 \\ f=\mp4$ . So (I'm tired at this point haha) $|a+b+c|=|0 \pm1 \pm3|=|\pm4|=4 \\ |d+e+f|=|\pm1+0\mp4|=|\mp3|=3$ and $|a+b+c| +|d+e+f|=4+3=\boxed{7}$

We find by inspection that $f(x,y) = (x+3)(y-4)$ . Thus we simply plug in and find $3+4 = 7$

It is a simple problem . JUST START CONSIDERING VALUES OR ASSUME.HERE I HAVE SIMPLY ASSUMED THAT a=1;b=0;c=3;d=0;e=1;f=-4 YOU FIND THIS BY SIMPLE ANALYSIS OF PROBLEM

First, we have that (ax+by+c)(dx+ey+f)=xy+3y-4x-12 (ad)x^{ 2 }+(be){ y }^{ 2 }+(ae+bd)xy+(af+cd)x+(bf+ce)y+cf=xy+3y-4x-12 So,it's sure that ad=0; be=0 and cf=-12. But as { a,b,c,d,e,f } are entiers, then c=3 and f=-4. So b and d are 1. Finally the absolue valeur is 7.

We can factorize the function to the form $(x+3)(y-4)$ . So we can conclude that $a=e=1, b=d=0, c=3$ and $f=-4$ . Thus, $|a+b+c|+|d+e+f|=|1+0+3|+|0+1+(-4)|=|4|+|-3|=4+3=7$

$xy+3y-4x-12=(x+3)(y-4)$

The expression is so easy to factorize.

\f(x,y)=(x+3)(y-4)]

$xy+3y-4x-12=(x+3)(y-4)$ This implies, $a=1,b=0,c=3,d=1,e=0,f=-4$ , so the answer is $2$

1. Group the terms so as to acquire common monomials. (xy + 3y) - (4x + 12)
2. Factor out common terms y(x+3) - 4(x + 3)
3. Combine the common terms and it will result to: (y-4) (x+3)

Add the coefficients and take the absolute values.

|1-3| + |1+3| = 7

We can factor out y in the first two terms and -4 in the last two terms, giving us y(x+3) - 4(x+3). From this, we can further factor out (x+3), giving us (x+3)(y-4). Therefore, if we evaluate the expression |a + b + c| + |d + e + f|, we get |1+3| + |1-4| = 7.

(ax + by + c)( dx + ey + f) = xy - 4x + 3y -12 ---> adx^2 + aexy + afx + bdxy + bey^2 + bfy + cdx + cey + cf = xy - 4x + 3y -12 ---> adx^2 + (ae + bd)xy + bey^2 + (af + cd)x + (bf +ce)y + cf = 0.x^2 + 1xy + 0.y^2 - 4x + 3y - 12. Now , we have ad = 0, be = 0 , cf = - 12, ae + bd = 1, af + cd = - 4, bf + ce = 3 ; So we have solutions that are a = 0 , b = 1 , c = -4 , d = 1 , e = 0, f = 3. So the result of |a + b + c| + |d + e + f| = |0 + 1 + (-4)| + |1 + 0 + 3| = |-3| + |4| = 3 + 4 = 7. Answer : 7