Solving inequalities, simplifying radicals, and factoring. (Pre calculus)

Solving inequalities, simplifying radicals, and factoring. (Pre calculus)

(Q.1) Solve for x in (x^3) - 5x > 4(x)^2 its a question in pre calculus
for dummies workbook, chapter 2. The answer says: then factor the
quadratic: x(x-5)(x+1)>0. Set your factors equal to 0 so you can find your
key points.When you have them, put these points on a number line and plug
in test numbers form each possible section to determine whether the factor
would be positive or negative. Then, given that you're looking for
positive solution, think about the possibilities: (+)(+)(+) = +, ++- = - ,
-+- = +, --- = -. Therefore, your solution is -1 < x < 0 or x > 5. so i
know how he got the x>5 but i don't get the -1 < x < 0 cuz it suppose to
be x > -1 and what these possibilities have to do with the solution ?
please explain to me in details.
(Q.2) solve for x in (x^5/3) - 6x = (x^4/3) same book. The answer says:
Next, factor out an x from each term: x(x^2/3 - x^1/3 - 6) = 0. The
resulting expression is similar to y^3(y^2 - y - 6), which factors into
y^3(y+2)(y-3).Similarly, you can factor x(x^2/3 - x^1/3 - 6) into x(x^1/3
+ 2)(x^1/3 - 3) = 0 I don't get how did he factor the main equation and
x^5/3 became x^2/3 and x^4/3 became x^1/3. I know how to factor like this
but with numbers not fractions. And also how this expression x(x^2/3 -
x^1/3 - 6) = 0 is similar to that y^3(y^2 - y - 6). I'm solving for x so
why he got the y into the answer now ?! (weird)
(Q.3) simplify 8/(4^2/3) same book. The answers says: change it to
8/sqrt[3]{4^2} ( i get this one ) but then he said multiply the numerator
and denominator by one more cube root of 4 . so how i can multiply 8/
\sqrt[3]{4^2} to 8/ \sqrt[3]{4} and get 4 ? isn't supposed to be 8/
\sqrt[3]{4^2} X 8/ \sqrt[3]{4^2} equals 4 ?