SOLUTION: This problem involves factoring terms that include positive and negative exponents. For some reason I just can't see it:
(3x+2)(2/3)(6x-5)^(-1/3)(6)+(6x-5)^(2/3)(3)
I keep co
Algebra ->
Exponents-negative-and-fractional
-> SOLUTION: This problem involves factoring terms that include positive and negative exponents. For some reason I just can't see it:
(3x+2)(2/3)(6x-5)^(-1/3)(6)+(6x-5)^(2/3)(3)
I keep co
Log On
Question 341826: This problem involves factoring terms that include positive and negative exponents. For some reason I just can't see it:
(3x+2)(2/3)(6x-5)^(-1/3)(6)+(6x-5)^(2/3)(3)
I keep coming up with:
I factored out (6x-5)^(-1/3)(2/3)(3)[(3x+2)+2(6x-5)^(3/3) <- (or 1)
So that simplifies to: (6x-5)^(-1/3)(2)(3x+2)+12x-10
Further simplified to: (6x-5)^(-1/3)[6x+4+12x-10]
The (6x-5)^(-1/3) moves to the denominator and 18x-6 remains in the numerator.
I don't know where I am going wrong with this. I know the denominator is right, but the answer shown for the numerator is 30x-7 and I just can't figure out how to get to that. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! (3x+2)(2/3)(6x-5)^(-1/3)(6)+(6x-5)^(2/3)(3)
---
[4(3x+2)*(6x-5)^(-1/3)] + [3(6x-5)^(2/3)]
---
Common Factor: (6x-5)^(-1/3)
---
= CF[12x+8 + 3(6x-5)]
---
= CF[12x+8+18x-15]
---
= CF[30x-7]
---
= [30x-7]/(6x-5)^(1/3)
===========================
Cheers,
Stan H.
I know the denominator is right, but the answer shown for the numerator is 30x-7 and I just can't figure out how to get to that.
(