How to eliminate negative exponents when finding critical points

[Quadruple Bypass]

OK, the professor did this problem for us as an example and i got lost somewhere in between. the problem was: find the critical points
f(x)= x^(4/5)*(x-4)^2

then he used the product rule to get

f '(x)= 4/5 x^(-1/5) *(x-4)^2 + x^(4/5) * 2(x-4) = 0

THEN, the part that threw me off was the next part where he said multiply both sides by 1/5...what did he mean by that? after that you get

4/5 (x-4)^2 + 2x(x-4) = 0 and so on...

my question is how and what did he do to get rid of the x^(-1/5) and the x^(4/5)?

Any help would be greatly appreciated

 

[NateTG]

Multiply both sides by
$$x^{\frac{1}{5}}$$

 

[shmoe]

He multiplied both sides by $$x^{1/5}$$, not 1/5.

 

[Hammie]

Don't multiply by 1/5, try multiplying by x^(1/5). The whole idea is to turn the exponents into integers, and then you can use the usual methods to find the roots of the equation.

 

[NonSequitur]

He is not multiplying by 1/5, he is multiplying by x^(1/5). On the zero side, it of course goes away. On the other side, it distributes and x^(1/5)*x^(-1/5)=1 while x^(1/5)*x^(4/5)=x nicely killing off those pesky rational exponents. Good Luck

 

[Quadruple Bypass]

WOOHOOO! Thanks very much!

 

[HallsofIvy]

"Never seen such unanimity of opinion before in my life"
Poobah, in "The Mikado"

 

[Hammie]

And who said mathmaticians don't have a sense of humor??