How do you solve x^3+4x^2>x+4 using a sign chart?

Nov 30, 2016

$- 4 < x < - 1$
$x > 1$

Explanation:

${x}^{3} + 4 {x}^{2} > x + 4$
${x}^{2} \left(x + 4\right) > x + 4$
${x}^{2} \left(x + 4\right) - \left(x + 4\right) > 0$
$\left({x}^{2} - 1\right) \left(x + 4\right) > 0$

We need to find the critical values (where a sign change can occur), which is given by $\left({x}^{2} - 1\right) \left(x + 4\right) = 0$

Either ${x}^{2} - 1 = 0 \implies x = \pm 1$
or $x + 4 = 0 \implies x = - 4$

So we need to examine the behaviour of (x^2-1)(x+4) in each of the intervals:

Sign of $\left({x}^{2} - 1\right) \left(x + 4\right) = \left\{\begin{matrix}\left(+\right) \left(-\right) < 0 & x < - 4 \\ \left(+\right) \left(+\right) > 0 & - 4 < x < - 1 \\ \left(-\right) \left(+\right) < 0 & - 1 < x < 1 \\ \left(+\right) \left(+\right) > 0 & x > 1\end{matrix}\right.$

So the solution for $\left({x}^{2} - 1\right) \left(x + 4\right) > 0$ is
$- 4 < x < - 1$ or $x > 1$