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

Aug 25, 2017

Solution : $- 0.4142 < x < 0 \ast \mathmr{and} \ast$ x >2.4142  **In interval notation:** (-0.4142,0) uu (2.4142,oo)

#### Explanation:

${x}^{3} > 2 {x}^{2} + x \mathmr{and} {x}^{3} - 2 {x}^{2} - x > 0 \mathmr{and} x \left({x}^{2} - 2 x - 1\right) > 0$ or

Roots of $\left({x}^{2} - 2 x - 1\right)$ are $x = \frac{2 \pm \sqrt{\left({2}^{2} - 4 \cdot 1 \cdot - 1\right)}}{2}$

or $x = 1 \pm \sqrt{2} \mathmr{and} x = 2.4142 , x = - 0.4142$

$\therefore x \left({x}^{2} - 2 x - 1\right) > 0 \mathmr{and} x \left(x - 2.4142\right) \left(x + 0.4142\right) > 0$

Critical points are $x = 0 , x = 2.4142 , x = - 0.4142$

Sign chart:

When $x < - 0.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is

 (-) * (-) * (-) = (-) ; <0 

When $- 0.4142 < x < 0$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is

 (-) * (-) * (+) = (+) ; >0 

When $0 < x < 2.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is

 (+) * (-) * (+) = (-) ; <0 

When $x > 2.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is

 (+) * (+) * (+) = (+) ; >0 

Solution: $- 0.4142 < x < 0 \mathmr{and}$ x >2.4142

In interval notation: $\left(- 0.4142 , 0\right) \cup \left(2.4142 , \infty\right)$

graph{x^3-2x^2-x [-10, 10, -5, 5]} [Ans]