Let $A=\{x| x^3-10x^2-20x-30>0\}$. Show $A$ is non empty and bounded below I have been trying to factor $x^3-10x^2-20x-30=0$ to find a solution because this would show $A$ is not empty and give me the lower bound I need but I can't solve for x. Any help is appreciated
 A: You need not find a solution of the equation. Just note that when $x$ is very large, the $x^3$ dominates the rest, so try something like $x=10$ to show that the LHS is positive.
A: To show is not empty is enough taking a $x_0$ such that $f(x_0)>0$. And clearly is not bounded (polynomial with positive leading coefficient). 
A: Look it as $x^3-10((x+1)^2+2)$, now $10((x+1)^2+1)$ is a parabola and after some $x=x_0$ it will be below the cubic $y=x^3$ because cube grows faster than quadratic and thus the set is non empty, also it is unbounded since for every value after the point $x=x_0$we must have  $x^3>10((x+1)^2+1)$.
A: Since 
$$ p(x)=x^3-10x^2-20x-30 $$
is a third degree polynomial, it has at least one real root $\xi$. It is a simple root since $p(x)$ and $p'(x)$ have no roots in common, so there is an element of $A$ in a right or left neighbourhood of $\xi$. $p(x)$ is a continuous function, hence $A$ is open and non-empty. Moreover, $A$ is bounded below since for any $x<-\sqrt{20}$ we have:
$$ p(x) = (x^2-20)x-(10x^2+30) < (x^2-20) x < 0.$$
