# How do you solve (x-4)^5(x-5)(x+3)<0 using a sign chart?

Jul 31, 2017

#### Answer:

$x \in \left(- \infty , - 3\right) \cup \left(4 , 5\right)$

#### Explanation:

Steps for solving using a sign chart:

1. Factor the inequality

2. Start with a relatively large value (greater than the largest zero) to determine where to start (positive or negative)

3. On a number line, draw the zeros of the factored inequality.

4. For each factor multiplied odd times, the curve should cross the number line at the zero. Otherwise, it touches the zero and bounces back.

5. Figure out the proper ranges that satisfy the inequality base on the abstract sketch of the inequality

We can directly start with the 2nd step here:

1. Evaluate the left-hand side with $x = 10$, and you get an apparently positive number. Therefore we should start from above the axis in the next step.

2. Draw a number line with all the zeros:
$x = - 3 \text{, "x=4", } x = 5$ (all of which should make the left-hand side $0$)
graph{(x+3)(x-4)(x-5) [-5,10, -10, 100]}

3. As you can see, all of the factors multiples odd times. Thus the line goes "penetrates" the line at all points.
(here's a graph for your reference)

4. The question requires negative values, so we take all segments under the number line (or $x$-axis, if you want,) that is, $x \in \left(- \infty , - 3\right) \cup \left(4 , 5\right)$

Jul 31, 2017

#### Answer:

The solution is x in (-oo,-3) uu (4,5))

#### Explanation:

Let

$f \left(x\right) = {\left(x - 4\right)}^{5} \left(x - 5\right) \left(x + 3\right)$

Let's build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$${\left(x - 4\right)}^{5}$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$ when x in (-oo,-3) uu (4,5))

graph{(x-4)^5(x-5)(x+3) [-18.02, 18.02, -9.01, 9.02]}