# How do you solve (x - 3) ( x - 4) ( x - 8) \leq 0?

Jun 15, 2018

$\left(x - 3\right) \left(x - 4\right) \left(x - 8\right) \le 0$ when $x \le 3$ and when $4 \le x \le 8$.

#### Explanation:

For this type of problem, it's probably easiest to draw a table like this:

Our expression is composed of three factors, $\left(x - 3\right)$, $\left(x - 4\right)$, and $\left(x - 8\right)$. These three factors together with their product serve as the top row heading for the table. The most left-hand column shows different possibilities for $x$, The next column to the right of this shows whether the value of $x - 3$ will be negative (-), zero (0), or positive (+) and the last column shows us if the product of the three factors is negative, zero, or positive.

Reading the first row below the heading we can see that when $x$ is less than 3, $x - 3$ is negative, $x - 4$ is negative, and $x - 8$ is negative, so the product of these three factors is also negative, since a negative times a negative times a negative is negative.

Looking at the table, we can see when the product of the factors is less than or equal to zero. This is when

$x \le 3$, and $4 \le x \le 8$.