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Mathematics 16 Online
OpenStudy (anonymous):

How do i factor this? what is the method to determine the factors 6b^2+4a-8b-3ab

OpenStudy (amistre64):

grouping seems most likely

OpenStudy (amistre64):

6b^2+4a-8b-3ab (-3ab-8b) + (6b^2+4a) -b(3a+8) + 2(3b^2 +2a)..... perhaps its: (6b^2-3ab)+(4a-8b) 3b(2b-a) +4(a-2b) thats better; now factor a -1 out of the first thing

OpenStudy (amistre64):

-3b(a-2b) +4(a-2b) = (a-2b)(4a-3b)

OpenStudy (amistre64):

(a-2b)(4-3b)...... typoed it :)

myininaya (myininaya):

i'm afraid he's one

myininaya (myininaya):

right*

OpenStudy (anonymous):

(4-3b) isnt one of my options, i have 4+3b

myininaya (myininaya):

any other options

OpenStudy (amistre64):

then they factored out a -1 from the other side itll look like this... 3b(2b-a) +4(a-2b) 3b(2b-a) -4(2b-a) (2b-a)(3b-4)......hmmm

OpenStudy (amistre64):

(2b-a)(3b+4)

OpenStudy (amistre64):

(2b-a)(4+3b)...... is also a way of writng it

OpenStudy (anonymous):

(8+3a), (2a+3b^2), (a-2b), (4+3b), (a+2b)

myininaya (myininaya):

or you could just multiply by (-1)(-1) (-1)(a-2b)(-1)(4-3b) (2b-a)(3b-4)

OpenStudy (amistre64):

(a-2b) is possible; and (4+3b) is possible; just not at the same time :)

OpenStudy (amistre64):

(2b-a)(4+3b) 8b +6b^2 -4a -3ab... which aint original; so that aint it

OpenStudy (amistre64):

(a-2b)(4-3b) 4a -3ab -8b +6b^2...is the original; so go with that. (a-2b)

OpenStudy (anonymous):

ok i will give that a shot. Thanks a lot

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