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nae
06-21-2006, 06:58 PM
please help me understand this?

12mb-9m+8b-6

I know i group them (4mb-3m) +(4b-3)

would it be 4m(b-3)?

if not could someone please help me understand.

galactus
06-21-2006, 07:49 PM
Hello nae:

The trick is to get what is inside the parentheses the same.

\L\\12mb-9m+8b-6

Factor:

\L\\3m(4b-3)+2(4b-3)

See there?. 4b-3 inside the parentheses. So, you have:

\L\\(3m+2)(4b-3)

tkhunny
06-21-2006, 07:49 PM
please help me understand this?

12mb-9m+8b-6

I know i group them (4mb-3m) +(4b-3)

would it be 4m(b-3)?How do you "know" you are to group them like that?

(4mb-3m) +(4b-3) = m*(4b-3) +(4b-3) = (m+1)(4b-3)

You can understand this better if you think about this

m*a + m = m*a + m*1 = m(a+1)

stapel
06-21-2006, 10:59 PM
I know i group them (4mb-3m) +(4b-3)
Yes; this was explained in your other thread (http://www.freemathhelp.com/forum/viewtopic.php?t=15222) on this exercise.

In the future, please post follow-ups and replies to the original threads. Thank you.

Eliz.