anonymous
  • anonymous
Using complete sentences, explain how to factor each one. Be sure that the final factorization (or "answer") is a part of your explanation. 2x^2 + 13x + 15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Help plzz!
anonymous
  • anonymous
Aren't you supposed to do this assignment yourself and then send it in? :P
anonymous
  • anonymous
@Nightie I dont think so. I just need help solving it.

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anonymous
  • anonymous
apoorvk will help you then ;)
anonymous
  • anonymous
ok. ?
anonymous
  • anonymous
He is typing a reply, I assume he is working on giving you a nice solution. Hence my previous response.
anonymous
  • anonymous
@Nightie yeah ot looks like that but i have another problem likethis one i need help on too but i am waiting on this one to be finished.
apoorvk
  • apoorvk
okay factoring a quadratic is done using what we call "splitting of the middle term". let the standard quadratic be ax^2 + bx + c. we basically take all the factors of 'a' and 'c' and rearrange them in two groups, so that subtracting or adding them i get 'b', taking care of the signs ofcourse. let me show you how. here we have ----> 2x^2 + 13x + 15 so 'a' and 'c' here are 2 and 15 combined prime factors are 1,2,3,5 so let me group (1,2,3), and (5). i get (1x2x3)+(5) = 6 + 5 =11. this is not equal to 'b' that is 13. so we try another combination. lets try for (1,2,5) and (3). here, (1x2x5)+(3)= 13 so this one will work!! so we split 13x as 10x+3x so now i write the expression as: 2x^2+13x+15 = 2x^2 + (10)x + (3)x + 15 = 2x(x+5) + 3(x+5) = (2x+3)(x+5) <------- There we are!! hope that cleared things up!!
anonymous
  • anonymous
@apoorvk ur an amzing person! thank you for all ur help.
anonymous
  • anonymous
Good job man!
apoorvk
  • apoorvk
Don't mention @suzan. that's what we are all here for right?
anonymous
  • anonymous
@apoorvk thats true, can u help me with one more? plz
apoorvk
  • apoorvk
yeah sure. btw, if its similar, why don't you try yourself for once, and post if you are getting stuck.
anonymous
  • anonymous
Ok, i will try it myself and if i get stuck i will post it.

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