Asked by Anonymous

I need help starting this, I don't want a plain answer just a what to do.

(bc'+a'd)(ab'+cd')-Simplify

My best guess is to use the distributive
property.

(a+b)(a+c)= a(b+c). But now that there is a 4th variable what do I do?



Answered by Steve
use it twice

(bc'+a'd)(ab'+cd')
= bc'(ab'+cd') + a'd(ab'+cd')
= abb'c' + bcc'd' + aa'b'd + a'cdd'

Now, since xx' = 0, the result is 0
Answered by Anonymous
OOK, thank you, I see how you did that step. that makes perfect sense.

so simplified you get

ac'+bd'+b'd+a'c
Answered by Steve
no, the result is 0, since

abb'c = a(bb')c = a0c = 0

each term is 0
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