Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

please factor 2a^2+30a+100

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

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

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

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.

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

hartnn (hartnn):

u can use http://www.wolframalpha.com/ for direct answers

OpenStudy (jiteshmeghwal9):

maths needs practice only & then there is no need of copying.

OpenStudy (anonymous):

shut up and thanks for the site hartnn

hartnn (hartnn):

@urjapatel plz, don't be rude, jitesh was trying to help u.

OpenStudy (anonymous):

what ever but thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!