<s>[INST] Finding real zeros of polynomial of third degree to solve inequality \begin{align}{}
  & \dfrac{x^3+2 x^2}{2} < x+2 \\[18pt]
  & x^3 + 2x^2 < 2x+4 \\
  & x^3 + 2x^2 -2x - 4 < 0 \\
\end{align}
Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$


*

*Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$

*Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$

*Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.
Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$
Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.
 [/INST] Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$


*

*Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$

*Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$

*Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.
Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$
Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.
</s>