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Mathematics
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OpenStudy (anonymous):
Will fan and give medals!
Factor 3y 2 - 4y.
y(4 - 3y)
y(3y - 4)
y(4y - 3)
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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?
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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?
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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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