SOLUTION: {cuberoot(128x^5)} I took the cube root and factored the 128 to get 2*64x^5 a 4 is pulled out and 2 is left in the cubed root. Im not sure what to do about the x^5 in the cube

Algebra ->  Radicals -> SOLUTION: {cuberoot(128x^5)} I took the cube root and factored the 128 to get 2*64x^5 a 4 is pulled out and 2 is left in the cubed root. Im not sure what to do about the x^5 in the cube      Log On


   

Question 887296: {cuberoot(128x^5)}
I took the cube root and factored the 128 to get 2*64x^5
a 4 is pulled out and 2 is left in the cubed root.
Im not sure what to do about the x^5 in the cubed root.
I think you pull out x^5 and get a finished product of:
(4x^5){cube root(2)}
Would this be correct? Thanks for the help!
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
{cuberoot(128x^5)}
root%283%2C128x%5E5%29
root%283%2C2%5E7x%5E5%29
2%5E2x%2Aroot%283%2C2x%5E2%29
4x%2Aroot%283%2C2x%5E2%29

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
{cuberoot(128x^5)}
I took the cube root and factored the 128 to get 2*64x^5
a 4 is pulled out and 2 is left in the cubed root.
Im not sure what to do about the x^5 in the cubed root.
I think you pull out x^5 and get a finished product of:
(4x^5){cube root(2)}
Would this be correct? Thanks for the help!


You need to break up the RADICAND into CUBES.

x%5E5+=+x%5E3%2Ax%5E2, so:

root%283%2C+128x%5E5%29 becomes: root%283%2C+4%5E3%2A2%2Ax%5E3x%5E2%29

root%283%2C+4%5E3%2Ax%5E3%2A2x%5E2%29 ------ Rearranging radicand

highlight_green%28highlight_green%284x%2Aroot%283%2C+2x%5E2%29%29%29 ----- Taking out 4 & x