# How do you solve and write the following in interval notation: x^3 - 38x^2 <0?

Aug 4, 2016

Solution is $x < 38$, except $x = 0$

#### Explanation:

${x}^{3} - 38 {x}^{2} < 0$ can be factorized as

${x}^{2} \left(x - 38\right) < 0$

Now, if $x < 0$, ${x}^{2} \left(x - 38\right) < 0$ as ${x}^{2}$ is positive, but $\left(x - 38\right)$ is negative and their product too is negative and inequality is satisfied.

If $0 < x < 38$, while ${x}^{2}$ is positive, $\left(x - 38\right)$ is negative and their product too is negative and inequality is satisfied.

But if $x > 38$, ${x}^{2}$ and $\left(x - 38\right)$ are positive, and their product too is positive, and inequality is not satisfied.

Hence, solution is $x < 38$, but $x = 0$ is not one as at $x = 0$, ${x}^{3} - 38 {x}^{2} < 0$ is not true.