# How do you solve x(3-x)(x-5)<=0?

Jul 31, 2016

Solution is $0 \le x \le 3$ or $x \le 5$.

#### Explanation:

As $x \left(3 - x\right) \left(x - 5\right) \le 0$, we have following options

either

(1) $x < 0$ - in such a case $x$ and $x - 5$ are negative and $3 - x$ is positive and hence $x \left(3 - x\right) \left(x - 5\right)$ is positive and hence $x < 0$ is not a solution.

(2) $0 \le x \le 3$ - in such a case $x$ and $3 - x$ are positive and $x - 5$ is negative and hence $x \left(3 - x\right) \left(x - 5\right)$ is negative and hence $0 \le x \le 3$ is a solution .

(3) $3 < x < 5$ - in such a case $x - 5$ and $3 - x$ are negative and $x$ is positive and hence $x \left(3 - x\right) \left(x - 5\right)$ is positive and hence $3 < x < 5$ is not a solution .

(4) $x \le 5$ - in such a case $x$ and $x - 5$ are positive and $3 - x$ is negative and hence $x \left(3 - x\right) \left(x - 5\right)$ is negative and hence $0 \le x \le 3$ is a solution.

Hence solution is $0 \le x \le 3$ or $x \le 5$.

graph{x(3-x)(x-5) [-10, 10, -10, 10]}