Question

completely factor: (3x+1)^6(2x-5)^(1/2)+(2x-5)^(3/2)(3x+1)^5

Answer 1

(2x-5)^1/2 (3x+1)^5 (5x-4)

Answer 2

um..could you explain please?

Answer 3

well i kinda did it the cheap way.. your going to keep your lowest powers (2x-5)^1/2 and (3x+1)^ 5

Then for the last bit you take 2x+3x and you get 5x and -5+4 = -1 soo you will get (5x-4)

Answer 4

-5+1=-4************

Answer 5

wt happened to the powers?

Answer 6

um....the powers??????

Answer 7

this one?

Answer 8

yep....

Answer 9

come over here and stay put so i can find you in the mathroom :)

Answer 10

oh...ok

Answer 11

there we go, i kept ending up on nomans land trying to get here on your profile page

Answer 12

is this what you wnat factored?
\[(3x+1)^6(2x-5)^{1/2}+(2x-5)^{3/2}(3x+1)^5\]

Answer 13

yea.

Answer 14

this thing is definitely very buggy and frustrating. - -

Answer 15

lets clean it up:

\[(A)^6(B)^{1/2}+(B)^{3/2}(A)^5\]
\[(A)^5\ ((A)(B)^{1/2}+(B)^{3/2})\]
\[(A)^5(B)^{1/2}\ ((A)+(B))\]
we good so far?

Answer 16

0.0 that's really amazing......

Answer 17

Which part, that i have the ability to factor? or that it cleans up when you rename stuff? :)

Answer 18

um.....both i guess?

Answer 19

lol .... you got the rest of it?

Answer 20

sooo....that's it? just substitue in the variables?

Answer 21

0.0 and thx for the medals :)

Answer 22

even thou i did nothin'