# How do you solve x^3+x^2+4x+4>0 using a sign chart?

Feb 4, 2017

The answer is x in ]-1, +oo[

#### Explanation:

Let $f \left(x\right) = {x}^{3} + {x}^{2} + 4 x + 4$

Before, we need to find the factors of $f \left(x\right)$

$f \left(- 1\right) = - 1 + 1 - 4 + 4 = 0$

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

To find the other factors, we perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} + {x}^{2} + 4 x + 4$$\textcolor{w h i t e}{a a a a}$$|$$x + 1$

$\textcolor{w h i t e}{a a a a}$${x}^{3} + {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$${x}^{2} + 4$

$\textcolor{w h i t e}{a a a a a}$$0 + 0 + 4 x + 4$

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

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

Therefore,

$f \left(x\right) = \left(x + 1\right) \left({x}^{2} + 4\right)$

$\forall x \in \mathbb{R} , \left({x}^{2} + 4\right) > 0$

So, we can 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}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

Therefore,

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