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O.G. 11

by vladmire » Thu Jan 29, 2009 7:02 am
Hi Question # 67. page 308.

Can some explain how I factor (1) ab + cd + ac + bd=6

I understand that like terms are grouped together i.e
(ab + bd) + (cd + ac)=6
I also understand that b and c can be factored out leaving
b(a + d) + c(a+d) = 6
my confusion lies in how under the explanation one of the (a+d) terms dissappears where does it go?
why is it not (b+c)(a+d)(a+d) = 6
please help me
even though technically you don't have to completely solve the problem...
Thanks in ADVANCe!!!
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by sanjay_dce » Thu Jan 29, 2009 8:14 am
it is as simple as follows. suppose xy -xz then x can be taken out common , so x(y-z) similarly take common term out and then see if you understand
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by DanaJ » Thu Jan 29, 2009 8:16 am
b(a + d) + c(a+d) = (b + c)(a + d), because you consider a + d to be the factor. Think of it this way: let's say a + d = x. Then you get that
b(a + d) + c(a+d) = bx + cx = x (b + c). Substitute x with a + d and you get that x (b + c) = (a + d)(b + c).
One of the (a + d) does not disappear, it's still there, represented by [b]c[/b].
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