# Basic algebra simplying/factoring question.

I went through basic algebra literally like 20+ years ago, I generally understand it well, but vaguely remember specifics if any.

Right now I am doing an advanced programming course and one question had an answer which did some basic algebraic simplification. I do not understand how this works anymore. Here is the steps of simplification it went through:

• Step $$1:$$ $$a \cdot b + a \cdot c - (b + c)$$
• Step $$2:$$ $$a \cdot (b + c) - (b + c)$$
• Step $$3:$$ $$(a - 1) \cdot (b + c)$$

I understand step 1 to step 2 fairly well. But how does step 2 end up as step 3?

Could someone point me to a video, tutorial or something that would refresh my understanding of whatever it is that allowed step 2 to get to step 3?

The idea between steps two and three is to factor out $$(b+c)$$.

This is easier for you to see if we replace $$(b+c)$$ by a single variable - let's have $$z = b+c$$. Then, starting at step two, we see

$$a\cdot (b+c) - (b+c) = a\cdot z - z$$

To be sure the next step is perfectly clear, don't forget that $$z = 1\cdot z$$, so we can say

$$a\cdot z - z = a\cdot z - 1\cdot z$$

Then you can factor out the $$z$$:

$$a\cdot z - 1\cdot z = z \cdot (a-1)$$

Then you replace $$z$$ back by $$b+c$$ again:

$$z \cdot (a-1) = (b+c) \cdot (a-1)$$

To get the exact representation in step four, you can just use the fact that multiplication commutes, i.e. $$x\cdot y = y\cdot x$$ (order doesn't matter). Thus,

$$(b+c) \cdot (a-1) = (a-1)\cdot (b+c)$$

Of course you can just leave it as $$(b+c) \cdot (a-1)$$, it's literally the same thing. Purely up to personal taste.

• The missing step which clarifies this is that $z$ can be written s $1\cdot z$ to have an explicit factor. Then $$a\cdot z-z=a\cdot z -1\cdot z=(a-1)\cdot z$$ – MPW Mar 8 '19 at 2:51
• Good point, I'll edit that in. – Eevee Trainer Mar 8 '19 at 2:51
• thanks that did make it perfectly clear again – rygo6 Mar 8 '19 at 2:55