# How do you solve 2x^3+5x^2>6x+9 using a sign chart?

Dec 30, 2016

The answer is x in ] -3, -1 [ uu ]3/2, +oo[

#### Explanation:

Let's rewrite the inequality as

$2 {x}^{3} + 5 {x}^{2} - 6 x - 9 > 0$

We must factorise the LHS

Let $f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} - 6 x - 9$

We calculate

$f \left(- 1\right) = - 2 + 5 + 6 - 9 = 0$

Therefore, $\left(x + 1\right)$ is a factor

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + 5 {x}^{2} - 6 x - 9$$\textcolor{w h i t e}{a a a a}$∣$x + 1$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$∣$2 {x}^{2} + 3 x - 9$

$\textcolor{w h i t e}{a a a a a a}$$0 + 3 {x}^{2} - 6 x$

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

$\textcolor{w h i t e}{a a a a a a a a a a}$$+ 0 - 9 x - 9$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$- 9 x - 9$

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

Therefore,

$\frac{2 {x}^{3} + 5 {x}^{2} - 6 x - 9}{x + 1} = 2 {x}^{2} + 3 x - 9 = \left(2 x - 3\right) \left(x + 3\right)$

So,

$2 {x}^{3} + 5 {x}^{2} - 6 x - 9 = \left(x + 1\right) \left(2 x - 3\right) \left(x + 3\right) > 0$

Now we can do the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$\frac{3}{2}$$\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 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}$$x + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{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}$$2 x - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{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}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{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}$$+$

Therefore,

$f \left(x\right) > 0$ when x in ] -3, -1 [ uu ]3/2, +oo[