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Mathematics 7 Online
OpenStudy (anonymous):

Will fan and give medals! Factor 3y 2 - 4y. y(4 - 3y) y(3y - 4) y(4y - 3)

OpenStudy (xapproachesinfinity):

what is the common factor here? any idea

OpenStudy (anonymous):

nope i have no idea. :/

OpenStudy (xapproachesinfinity):

we have y shared btw both terms yes

OpenStudy (anonymous):

huh?

OpenStudy (xapproachesinfinity):

let me rewrite it this way \[3y^2-4y=3yy-4y\] do you see that y is common?

OpenStudy (xapproachesinfinity):

say yes no, get out of here just say something lol

OpenStudy (xapproachesinfinity):

3y^2 is the same is as 3y times y yes?

OpenStudy (anonymous):

yes i did

OpenStudy (xapproachesinfinity):

okay good!

OpenStudy (anonymous):

so you will it b the second one?

OpenStudy (xapproachesinfinity):

now if y is common we can pull it out using distributive property \[3yy-4y=y(3y-4)\]

OpenStudy (xapproachesinfinity):

turns out to be B yes

OpenStudy (xapproachesinfinity):

if you did distributive property you will end up with what you started with 3y^2-4y that's why i said we are using that arithmetic law to get this

OpenStudy (xapproachesinfinity):

a(b+c)=ab+ac right?

OpenStudy (xapproachesinfinity):

so walking backwards is what is called factorization they give us ab+ac want us to get a(b+c)

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