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

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 But my math is a bit outdated and I cannot remember the law to see this.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.

• You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$ – Dashi Nov 26 '18 at 18:42
• Your title has $+$ instead of $*$ – Ross Millikan Nov 26 '18 at 18:54
• aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!! – InDaPond Nov 26 '18 at 18:57

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