Which mathematical law is used in $ab+ac-(b+c)=(a-1)(b+c)$

Question

I just stumbled upon a question to figure out how to simplify

J = (ab)+(ac)-(b+c)

My steps:

<=> a*(b+c)-b-c

<=> a*(b+c) -1*(b+c)

But that was not one of the solutions. One of these was, as mentioned above, (a-1)*(b+c).

As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.

Answer 1

It is called the distributive law $$ (x+y)z=xz+yz. $$ You apply it backwards in $$ (a-1)\underbrace{(b+c)}_{z}=a\underbrace{(b+c)}_{z}-1\underbrace{(b+c)}_{z}. $$

Answer 2

The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.