A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
factor: 6t^2t15
anonymous
 3 years ago
factor: 6t^2t15

This Question is Closed

DDCamp
 3 years ago
Best ResponseYou've already chosen the best response.0No, that would multiply to 7t²  28t. Have you learned the quadratic equation yet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ladyrosebud how do you figure it out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah @ladyrosebud is correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I made an error :( its 7t^228t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, thats what I thought

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We can do this using a method called "Product  Sum" Method The standard form of a Quadratic Function is \[Ax ^{2} + Bx + C\] We have to find two numbers such that their Product is A.C and their Sum is B In the given function \[6t ^{2}t15\] A = 6; B = 1 and C = 15 I have to find two numbers that I multiply should give me 90 and add them should give me 1. So Product is 6 x 15 = 90 Sum is 1 dw:1368671725295:dw Now set up a table like this and list all product pairs for 90.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1368671986360:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now my product is negative so one number should be positive and the other number should be negative. now when I look at all the product pair \[\pm 9 \space and \space \pm 10\] will give me 1 if '1' if I add them one being negative and one being positive. dw:1368672235371:dw So the numbers are either +9 and 10 or 9 and +10. Since the sum is 1 the numbers are +9 and 10.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since my A is not equal to 1 I have to rewrite the function with these two numbers. \[6t ^{2}  t 15 = 6t ^{2} + 9t + 10t  15\] Now I group them. \[(6t ^{2} + 9t ) + ( 10t  15 )\] Now the GCF of the first group is 3t if I take that out, it will be \[3t(2t + 3) + (10t 15)\] Now I have to take a GCF out of the second group such that I get a (2t + 3) with them too. so if I take a '5' out I will get a 2t + 3 . So I do\[3t(2t + 3) + 5 (2t + 3) = 3t (2t + 3) 5 (2t + 3)\] Now the common factor in both terms is (2t + 3), Doing distributive property in reverse, you will get \[(2t + 3) ( 3t  5 )\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer (2t+3)(3t5) was answered wayyy long ago. factoring this isn't that complexed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Gummy asked how did lady figured it out? that is why I tried explaining my bad.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol my bad i just kept seeing people solving and solving for the expression so i was like it shouldn't take this long
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.