# How do you solve and write the following in interval notation: x³ - 36x ≤ 0?

Jun 10, 2016

Solution set is $x \le - 6$ and $0 \le x \le 6$

#### Explanation:

Before we try to solve the inequality ${x}^{3} - 36 x \le 0$, let us factorize ${x}^{3} - 36 x$.

${x}^{3} - 36 x = x \left({x}^{2} - 36\right) = x \left(x - 6\right) \left(x + 6\right)$

We may write this as $\left(x + 6\right) x \left(x - 6\right)$

Hence, the term ${x}^{3} - 36 x$ is a multiplication of $\left(x + 6\right)$, $x$ and $\left(x - 6\right)$ and for equality part, we have solutions $x = - 6$, $x = 0$ and $x = 6$.

These three points divide number line in four parts.

(1) In segment $x < - 6$, all the three terms of are negative and hence product is negative i.e. less than zero and this is a solution.

(2) In segment $- 6 < x < 0$, while first term $\left(x + 6\right)$ is positive, other two terms are negative. Hence product is positive and this is not a solution.

(3) In segment $0 < x < 6$ while first two terms are positive, the third term $\left(x - 6\right)$ is negative, hence product is negative and this is a solution.

(4) In segment $x > 6$ all the terms are positive, hence product is positive and hence this is not a solution.

Hence solution set is $x \le - 6$ and $0 \le x \le 6$