# How do you solve 2x^3-3x^2-32x+48>=0 using a sign chart?

Jan 15, 2017

The answer is x in [-4,3/2] uu [4, +oo[

#### Explanation:

Let $f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 32 x + 48$

Then

$f \left(4\right) = 128 - 48 - 128 + 48 = 0$

So, $\left(x - 4\right)$ is a factor of $f \left(x\right)$

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 3 {x}^{2} - 32 x + 48$$\textcolor{w h i t e}{a a a a}$∣$\textcolor{red}{x - 4}$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 8 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$∣$\textcolor{b l u e}{2 {x}^{2} + 5 x - 12}$

$\textcolor{w h i t e}{a a a a a a}$$0 + 5 {x}^{2} - 32 x$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 5 {x}^{2} - 20 x$

$\textcolor{w h i t e}{a a a a a a a a a a}$$+ 0 - 12 x + 48$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$- 12 x + 48$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$- 0 + 0$

Therefore,

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 32 x + 48 = \left(x - 4\right) \left(2 {x}^{2} + 5 x - 12\right)$

$= \left(x - 4\right) \left(2 x - 3\right) \left(x + 4\right)$

Now, we can make 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 a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$\frac{3}{2}$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\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 a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 x - 3$$\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 a}$$+$$\textcolor{w h i t e}{a a a}$$+$

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

Therefore,

$f \left(x\right) \ge 0$ when  x in [-4,3/2] uu [4, +oo[