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Mathematics 18 Online
OpenStudy (anonymous):

Factory 12a^2 + 12ab+ 3b^2

OpenStudy (across):

Sorry, I've never heard of a factory with such name. On a totally unrelated note, if you want to factor \(12a^2+12ab+3b^2\), then you could observe that \(3(2a+b)^2\) will yield the same expression. Sorry for going a little off topic with that paragraph above.\[\]

myininaya (myininaya):

She knows.

OpenStudy (anonymous):

whats the confusion about then :o

OpenStudy (mr.math):

lol, across!

myininaya (myininaya):

I can show you another way to factored this without using trial factors if you like.

myininaya (myininaya):

She didn't know what you meant about the term factory. There is a mathematical term factor

myininaya (myininaya):

Well she did know.

myininaya (myininaya):

She was just making a little joke.

OpenStudy (anonymous):

nd sure.. i just hate factoring polynomials

myininaya (myininaya):

Factory is not a mathematical term.

OpenStudy (anonymous):

OH WOW it was a typo my bad

OpenStudy (anonymous):

LOL haha

myininaya (myininaya):

lol

OpenStudy (anonymous):

anyway can u help me through this ?

myininaya (myininaya):

Sure.

OpenStudy (anonymous):

myin lies :P http://mathfactory.org/ is completely mathematics :D

OpenStudy (anonymous):

Lol

myininaya (myininaya):

12a^2 + 12ab+ 3b^2 So we have this and this means \[12a^2+12ab+3b^2\] The GCF of 12,12,3 is 3 So we have \[3(4a^2+4ab+b^2)\] Now lets just look at the inside which is \[4a^2+4ab+b^2 \text{ note : we will need \to bring that 3 down later }\] \[4\cdot a^2+4b \cdot a+b^2\] The trick is to find two factors of A*C that have product A*C and have sum B. A=4 , B=4b , C=b^2 A*C=4b^2=2b*2b and B=4b=2b+2b So now remember our middle term was 4b*a and we have above 4b=2b+2b so 4b*a=(2b+2b)*a=2ab+2ab So we need to replace our middle term 4ab with 2ab + 2ab and then factor by grouping \[4a^2+2ab+2ab+b^2\] The first two terms have 2a in common so we have \[2a(2a+b)+2ab+b^2\] The last two terms have b in common so we have \[2a(2a+b)+b(2a+b)\] Now we have two terms and they both have (2a+b) in common so we can factor that out leaving us with \[(2a+b)(2a+b)\] We can also write this as \[(2a+b)^2\] Don't forget about the 3 we were suppose to bring down. \[3(2a+b)^2\]

OpenStudy (anonymous):

thanks this is so much better then just getting the answer

myininaya (myininaya):

Well with experience some of these will become easier for you to just give the answer.

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