## gumstopper99 2 years ago factor: 6t^2-t-15

1. gumstopper99

Oops . 7t(t-4) ??

Hi,batman.

3. DDCamp

No, that would multiply to 7t² - 28t. Have you learned the quadratic equation yet?

(2t+3)(3t-5)

5. gumstopper99

@ladyrosebud how do you figure it out?

6. galacticwavesXX

7. gumstopper99

I made an error :( its 7t^2-28t

7t(t-4)

9. rajee_sam

Ok can I help?

10. gumstopper99

Yes, thats what I thought

11. kyusakazaki

it's (t-5/3)(t+3/2)

12. 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.

13. rajee_sam

|dw:1368671986360:dw|

14. 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.

15. 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 $(2t + 3) ( 3t - 5 )$

16. galacticwavesXX

the answer (2t+3)(3t-5) was answered wayyy long ago. factoring this isn't that complexed

17. rajee_sam