anonymous
  • anonymous
factor: 6t^2-t-15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Oops . 7t(t-4) ??
anonymous
  • anonymous
Hi,batman.
DDCamp
  • DDCamp
No, that would multiply to 7t² - 28t. Have you learned the quadratic equation yet?

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anonymous
  • anonymous
(2t+3)(3t-5)
anonymous
  • anonymous
@ladyrosebud how do you figure it out?
anonymous
  • anonymous
yeah @ladyrosebud is correct
anonymous
  • anonymous
I made an error :( its 7t^2-28t
anonymous
  • anonymous
7t(t-4)
rajee_sam
  • rajee_sam
Ok can I help?
anonymous
  • anonymous
Yes, thats what I thought
anonymous
  • anonymous
it's (t-5/3)(t+3/2)
rajee_sam
  • rajee_sam
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 \[6t ^{2}-t-15\] 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.
rajee_sam
  • rajee_sam
|dw:1368671986360:dw|
rajee_sam
  • rajee_sam
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.
rajee_sam
  • rajee_sam
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 \[(2t + 3) ( 3t - 5 )\]
anonymous
  • anonymous
the answer (2t+3)(3t-5) was answered wayyy long ago. factoring this isn't that complexed
rajee_sam
  • rajee_sam
Gummy asked how did lady figured it out? that is why I tried explaining my bad.
anonymous
  • anonymous
lol my bad i just kept seeing people solving and solving for the expression so i was like it shouldn't take this long

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