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

(Q.1) Solve for $x$ in $x^3 - 5x > 4x^2$ 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 $x^\frac{5}{3} - 6x = x^\frac{4}{3}$ same book. The answer says: Next, factor out an $x$ from each term: $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$. 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 $x(x^\frac{2}{3} - x^\frac{1}{3} - 6)$ into $x(x^\frac{1}{3} + 2)(x^\frac{1}{3} - 3) = 0$ I don't get how did he factor the main equation and $x^\frac{5}{3}$ became $x^\frac{2}{3}$ and $x^\frac{4}{3}$ became $x^\frac{1}{3}$. I know how to factor like this but with numbers not fractions. And also how this expression $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$ 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 $\frac{8}{4^\frac{2}{3}}$ same book. The answers says: change it to $\frac{8}{\sqrt[3]{4^2}}$ ( 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 $\frac{8}{\sqrt[3]{4^2}}$ to $\frac{8}{ \sqrt[3]{4}}$ and get $4$ ? isn't supposed to be $\frac{8}{\sqrt[3]{4^2}}*\frac{8}{\sqrt[3]{4^2}}$equals $4$ ?

• math.stackexchange.com/a/441024/26947 I think you mean this one. – eccstartup Aug 20 '13 at 6:37
• Thanks for the editing but this is not the one. – Out Of Bounds Aug 20 '13 at 7:10
• The table is somewhat the same as the picture in the link I have given. For polynomials, they go to infinity at infinity points. And changes signs(+/-) at zero points with odd degree. That explains the picture. – eccstartup Aug 20 '13 at 7:26

Just for the first Question, as you're given, set $p(x)=x^3-4x^2-5x$. We are asked to make $p(x)$ positive, so one can factor it to its basic parts as $$p(x)=x(x-5)(x+1)$$ and then follow the table below:

I think you can easily find out why that interval is taken. (-:

• I'm sorry but i don't get the table ! can you explain further ? – Out Of Bounds Aug 20 '13 at 7:09
• @Amr: We allocate a row for each part "$x$" "$x-5$" and "$x+1$". Moreover, we allocate a row at the final for "$p(x)$". At the top of the table we write some things such that "$x$", "-infinity", "the roots of parts from small to big" and "+infty" at last. You have some boxes below as you can see how I drew them. Each root is associated to one part below. For example, we make that small segment below $0$ which is considered as a root for "$x$". The sign of "$x$" in the first row is $+1$. – mrs Aug 20 '13 at 7:18
• @Amr: So, we point the opposite sign $-$ before $0$ and the same sign $+$ after that. Got it?? (-: – mrs Aug 20 '13 at 7:20
• Okay got it. can you explain question 3 please too ? – Out Of Bounds Aug 20 '13 at 8:22
• I understand completely, about the need for a blackboard!! :-) – Namaste Aug 20 '13 at 11:43

multiply the numerator AND denominator by one more cube root of 4.

$$\frac{8}{4^{\frac{2}{3}}} = \frac{8}{\sqrt[3]{4^2}}\cdot \frac{\sqrt[3]{4^1}}{\sqrt[3]{4^1}} = \frac{8\cdot\sqrt[3]4}{\sqrt[3]{4^3}}=\frac{8\cdot\sqrt[3]4}{4}=2\cdot\sqrt[3]4$$