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gumstopper99Best ResponseYou've already chosen the best response.0
Oops . 7t(t4) ??
 11 months ago

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

gumstopper99Best ResponseYou've already chosen the best response.0
@ladyrosebud how do you figure it out?
 11 months ago

galacticwavesXXBest ResponseYou've already chosen the best response.0
yeah @ladyrosebud is correct
 11 months ago

gumstopper99Best ResponseYou've already chosen the best response.0
I made an error :( its 7t^228t
 11 months ago

gumstopper99Best ResponseYou've already chosen the best response.0
Yes, thats what I thought
 11 months ago

kyusakazakiBest ResponseYou've already chosen the best response.0
it's (t5/3)(t+3/2)
 11 months ago

rajee_samBest ResponseYou've already chosen the best response.0
We 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.
 11 months ago

rajee_samBest ResponseYou've already chosen the best response.0
dw:1368671986360:dw
 11 months ago

rajee_samBest ResponseYou've already chosen the best response.0
Now 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.
 11 months ago

rajee_samBest ResponseYou've already chosen the best response.0
Since 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 \[(2t + 3) ( 3t  5 )\]
 11 months ago

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

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

galacticwavesXXBest ResponseYou've already chosen the best response.0
lol my bad i just kept seeing people solving and solving for the expression so i was like it shouldn't take this long
 11 months ago
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