Question

factor completely and show steps:
2((x+4)^1/2)(2)((3x-2)^3)+((x+4)^-1/2)((3x-2)^4)(1/3)

Answer 1

Uh, can you show a picture of the equation or something? That looks suspicious...

Answer 2

\[2(x+4)^{1/2}(2)(3x-2)^{3}+(x+4)^{-1/2}(3x-2)^{4}(1/3)\]

Answer 3

Well, you could put that over a common denominator by multiplying the left hand term by \[\frac{3\sqrt{x+4}}{3\sqrt{x+4}}\] and then factor the numerator, but I'm still a bit skeptical.

Answer 4

why would you make it a fraction if it isn't one?

Answer 5

so you can combine the two terms into one.

Answer 6

like \[1+\frac{1}{2} = \frac{3}2\]

Answer 7

but couldn't you factor out the (x+4)^1/2 and the (3x-2)^3?

Answer 8

And get what?

Answer 9

Try it your way, try it mine, pick whichever one gets you a nicer looking answer...I gotta hit the sack.

Answer 10

\[2(x+4)^{1/2}(2)(3x-2)^{3} + (x+4)^{-1/2}(3x-2)^{4}(1/3) \]
\[(3x-2)^{3}[2(x+4)^{1/2}(2) + (x+4)^{-1/2}(3x-2)(1/3)]\]
\[(3x-2)^{3}[4(x+4)^{1/2} + (1/3)(x+4)^{-1/2}(3x-2)]\]
\[(3x-2)^{3}[4(x+4)^{1/2} + \frac{ 3x-2 }{ 3(x+4)^{1/2} }]\]
\[(3x-2)^{3}[\frac{ 12(x+4) }{ 3(x+4)^{1/2} } + \frac{ 3x-2 }{ 3(x+4)^{1/2} }] \]
\[(3x-2)^{3}[\frac{ 12x + 48 + 3x - 2 }{ 3(x+4)^{1/2} }] \]
\[(3x-2)^{3}[\frac{ 15x + 46 }{ 3(x+4)^{1/2} }] \]
\[\frac{ (15x + 46)(3x - 2)^{3} }{ 3(x + 4)^{1/2} }\]

Answer 11

@cebroski gets exactly the answer you would had you put them over the common denominator

\[2\sqrt{(x+4)}*2(3x-2)^3 + \frac{(3x-2)^4}{3\sqrt{(x+4)}}\]Multiply left term by \[\frac{3\sqrt{x+4}}{3\sqrt{x+4}}\]to get\[
\frac{3\sqrt{x+4}}{3\sqrt{x+4}}*2\sqrt{(x+4)}*2(3x-2)^3 + \frac{(3x-2)^4}{3\sqrt{(x+4)}}\]\[=\frac{2*3*\sqrt{x+4}*\sqrt{x+4}*2*(3x-2)^3+(3x-2)^4}{3\sqrt{x+4}}\]\[=\frac{12(x+4)(3x-2)^3+(3x-2)^4}{3\sqrt{x+4}}\]\[=\frac{(3x+2)^3(12(x+4)+(3x-2)}{3\sqrt{x+4}} \]\[=\frac{(3x+2)^3(12x+48+3x-2)}{3\sqrt{x+4}} = \frac{(3x+2)^3(15x+46)}{3\sqrt{x+4}}\]

Answer 12

Rationalized, that would be\[\frac{(3x-2)^3(15x+46)}{3\sqrt{x+4}}*\frac{\sqrt{x+4}}{\sqrt{x+4}} = \frac{(3x-2)^3(15x+46)(x+4)^{1/2} } {3(x+4) }\]

(I see I accidentally changed the sign in the \((3x-2)^3\) to \((3x+2)^3\) in the last two lines of the previous post, sorry for any confusion)