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hartnn (hartnn):
first factor out 2, what u get ?
Parth (parthkohli):
A good start is factoring the 2 out.
OpenStudy (anonymous):
just 2
hartnn (hartnn):
yes
OpenStudy (anonymous):
factor out the 2
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OpenStudy (anonymous):
a is what i got
Parth (parthkohli):
??
Parth (parthkohli):
No.
OpenStudy (anonymous):
i dont know my stupid teacher wont tell me how to do it
OpenStudy (anonymous):
wait is it a^2
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hartnn (hartnn):
ok, what u get when u divide 30 and 100 by 2 ?
Parth (parthkohli):
... no.
OpenStudy (anonymous):
15 and 50
hartnn (hartnn):
so factoring 2 from 2a^2+30a+100 will give you 2(a^2+15a+50)
got this ?
OpenStudy (anonymous):
no i dont understand that
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hartnn (hartnn):
ok, \((AB+AC+AD)= A(B+C+D)\)
this is distributive property, got this ?
similarly \(2a^2+2*15a+2*50= 2(a^2+15a+50)\)
now ?
OpenStudy (anonymous):
(2a+10)(a+10) ???
OpenStudy (anonymous):
can you just give me the answer please
hartnn (hartnn):
nopes, this is not a answering site, this is learning site..
hartnn (hartnn):
if u want to learn, i can help u.
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OpenStudy (anonymous):
yeah but ur confusing me
hartnn (hartnn):
sorry for that.
OpenStudy (jiteshmeghwal9):
Assume that u have an equation \(\Large{ax^2+bx+c}\)
Break the middle term \(bx\) such that its product becomes the product of \(ax^2\) & \(c\) & sum becomes \(bx\).
or u can use the identity\[\Huge{x^2+(a+b)x+ab.}\]
OpenStudy (anonymous):
its okay
OpenStudy (anonymous):
forget it I'll jus copy someones answer that won't ind giving me the answer
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