anonymous
  • anonymous
ab + 2b +3a + 6 uhm grouping a term i thinkkkk? (:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I can't belive I forgot how to do this...thinking...
Mertsj
  • Mertsj
Factor b out of the first two terms and then factor 3 out of the last two terms.
anonymous
  • anonymous
Then you will end up b(a+2)3(a+2)

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anonymous
  • anonymous
lol you sure?
Mertsj
  • Mertsj
b(a+2)+3(a+2)
anonymous
  • anonymous
Yes thanks to Mertsj
Mertsj
  • Mertsj
Now you have two terms and they have the common factor a+2 so factor out a+2
anonymous
  • anonymous
Then don't you put the b and 3 together to become (b+3)(a+2)?
Mertsj
  • Mertsj
yes
anonymous
  • anonymous
So the answer is (b+3)(a+2)
anonymous
  • anonymous
Do you understand why? :)
anonymous
  • anonymous
uhm , a little i see how you get it . but i feel so dumb . -_- i have a test tomorrow and i'm scared i'm going to fail!
anonymous
  • anonymous
No!!! Don't feel dumb! Sometimes school is difficult but God can help you with the test. I will pray do you do good. :) Anymore questions? :)
anonymous
  • anonymous
Then try factoring this (x^2-10x+25)
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
Lol! I guess they left. :D
anonymous
  • anonymous
:P
anonymous
  • anonymous
O well see ya. :D
anonymous
  • anonymous
im hereeee
anonymous
  • anonymous
im trying to figure it out haha
anonymous
  • anonymous
Oh..!! Lol! Sorry! :D
anonymous
  • anonymous
(x-5)(x-5) ?
anonymous
  • anonymous
Well it isn't that hard you know It just needs lots of pratice
anonymous
  • anonymous
Your missing something...
anonymous
  • anonymous
It is correct but not complete
anonymous
  • anonymous
(x-5)+(x-5) ?
anonymous
  • anonymous
Nope
anonymous
  • anonymous
uhhmmmmmmmmmmmmmm i don't know :c
anonymous
  • anonymous
Give up?
anonymous
  • anonymous
lol yes .
anonymous
  • anonymous
It is (x-5)^2
anonymous
  • anonymous
how!?
anonymous
  • anonymous
Well both are commons
anonymous
  • anonymous
oh that's truee. here i have another question please come help?
anonymous
  • anonymous
ok

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