Question

how do you factor 9a^2+4bc-4c^2-b^2? PLEASE?!?!?!?!??!

Answer 1

Not much you can do with that..

Answer 2

um. yea. thanks

Answer 3

There is some partial factoring you can do if you separate out some differences of squares, e.g.

Answer 4

Try this and see where it goes:

\[9a^2-(b^2-4bc+4c^2)\]

Answer 5

i did that, and i got 3a-(b^2-4bc+4c^2) and now i have no idea what to do

Answer 6

\[\rightarrow 9a^2-(b-2c)^2\]

Answer 7

What happened to the other 3a in 9a^2?

Answer 8

OH!!!!!! so you turn it into a correct trinomial by changing the addition sign to a subtraction sign in front of the parenthesis, changing the signs of all of the terms! then you factor that correct trinomial to get (3a-b-2c)(3a-b-2c)

Answer 9

+2c at the end, otherwise, bingo!

Answer 10

oh. thanksss!!!!!!!!!! :D

Answer 11

I retract my earlier statement about there not being much to do; took me a minute to see all those squares.

Answer 12

haha ok(:

Answer 13

(3a+b-2c)(3a-b-2c)

Answer 14

?

Answer 15

Oops, (3a+b-2c)(3a-b+2c)

Answer 16

? Did I not distribute the negative sign correctly?

Answer 17

Yeah, going from here:
(3a+(b-2c))(3a-(b-2c)) you should get (3a+b-2c)(3a-b+2c)

Answer 18

oh, yes, and +b at the beginning. That's what I got too. I didn't catch that error in Hannah's

Answer 19

oh yes @jabberwock hes right, because a -b times a -b equals positive b^2

Answer 20

But you're using x^2-y^2=(x+y)(x-y)

Answer 21

wait, now im confusedddd..........so the original problem is 9a+(4bc-4c^2-b^2) right? then i changed the positive in front of the parenthesis to a negative. i get that. but i dont understand why you would change that. but we did anyways. so then i got 9a-(b^2-4bc+4c^2) because all the signs would be opposite. then we factored that and got (3a+b-2c)(3a-b+2c). I have no idea why. care to explain??

Answer 22

(3a+b-2c)(3a-b+2c) is correct.
\[\rightarrow 9a^2-3ab+6ac+3ab-b^2+2bc-6ac+2bc-4c^2\]

Answer 23

\[9a^2-(b^2-4bc+4c^2)\]\[9a^2-(b-2c)^2\]
Good to there?

Answer 24

@jabberwock yes, but i dont understand why you would change that positive in front of the parenthesis to a negative

Answer 25

Isn't that how it appeared in the original problem?

Answer 26

Oh...it's not written that way in what you wrote...

Answer 27

nope! in the original it was 9aPLUS(4bc-4c^2-b^2)

Answer 28

so why would you change that plus to a minus?

Answer 29

Ah, you're right. CliffSedge and I were both thinking it was (-)...
\[9a^2+(b-2c)^2\]is as far as you can go.

Answer 30

how?

Answer 31

OH, nevermind again. Heh, sorry.
Original problem:
\[9a^2+4bc-4c^2-b^2\]
Rearrange the terms:
\[9a^2-b^2+4bc-4c^2\]

Answer 32

ok,

Answer 33

Notice:
\[-b^2+4bc-4c^2\]is the same as\[-(b^2-4bc+4c^2)\]by factoring out -1

Answer 34

ok,

Answer 35

oh, so you just put a parenthesis around the trinomial, and distributed the -1 and changed all of the signs.!

Answer 36

Exactly :)

Answer 37

ok then you factor to (3a-b+2c)(3a-b+2c)!!!~

Answer 38

Right :) :)

Answer 39

No, wait

Answer 40

Not right

Answer 41

thanks so much!!!!!!!---what?? lol

Answer 42

\[9a^2-(b-2c)^2\]
Notice that you have two squared terms. You have
\[(3a)^2-(b-2c)^2\]
Yes?

Answer 43

yup

Answer 44

So this is called the "difference of two squares," which happens whenever you have a squared term minus another squared term. It ALWAYS factors like this:
\[m^2-n^2=(m+n)(m-n)\]

Answer 45

yes i learned that,

Answer 46

K, so in this case

m = 3a
n = 2b-c

What does it look like when you substitute?

Answer 47

(3a+2b-c)(3a-2b-c)

Answer 48

don't you mean n=b-2c?

Answer 49

Oops for like the 4th time...lol, sorry

Almost.
\[(3a+(b-2c))(3a-(b-2c))\]
Look at the -(b - 2c) and distribute the negative to get
-b + 2c

Answer 50

OK!!!!!! haha lol(: thankkkssss:D

Answer 51

Sure, sorry about all the mistakes.

Answer 52

no prob(: