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.

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

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