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I want to solve this problem:

$$(x . y . z + x . y + x)$$

Which turns into this when you group $x$

$$x . ( yz + y + 1 ) $$

What I don't understand is why is there a "1" at the end? Does the last $x $ turn into a 1? If so, why?

My math book is unclear, I'm having a hard time.

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    $\begingroup$ Hi @sinan-samet, have you consider that the result x.(yz+y+1) stems from a basic factorization, simply...? If you take $(a∗a+b∗a)$ and factorize it with a, you will have $a(a+b)$ $\endgroup$ – Andy K Jun 2 '16 at 10:24
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If you look at the equation,

$x.y.z+x.y+x$

$=x.y.z+x.y+x.1$

This is basic boolean algebra, $[a.1=a]$ $[a+1=1]$

So if you group $x$,$1$ will be left at the end,as boolean algebra follow Distributive law.

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  • $\begingroup$ Thanks a lot I had solved this problem later on but this makes it more clear! $\endgroup$ – Sinan Samet Jun 7 '16 at 11:09
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You should use the identity rule for the last x before using the distributive rule.

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