Find a real numbers $u,v$ such that $p(u)<0$ and $p(v)>0$

Question

Consider $p(x)=x^3+ax^2+bx+c$ where are $a,b$ and $c$ are real number .Find a real numbers $u,v$ such that $p(u)<0$ and $p(v)>0$

since this odd degree polynomial so it has at least one root we cal prove by intermediate value theorem

so this function $p(x)$ at least on real root

but how to find $u,v$ such that $p(u)<0$ and $p(v)>0$

Thank you.....

Answer 1

Suppose $u > 0$ and $u > -3a$, $u^2 > -3b$ and $u^3 > -3c$. Then we have $p(u) > 0$.
Similarly suppose $v < 0$ and $v < -3a$, $v^2 > 3b$ and $v^3 < -3c$. Then we have $p(v) < 0$.

Answer 2

Hint: Note that \begin{align} p(x)= x^3+ax^2+bx+c = x^3\left( 1+\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}\right) \approx x^3 \end{align} when $|x|\gg 1$.