How do you solve x^3+5x^2-4x-20>=0 using a sign chart?

Oct 17, 2017

Solution: $- 5 \le x \le - 2 \mathmr{and} x > 2$ . In interval notation:
$x | \left[- 5 , - 2\right] \cup \left[2 , \infty\right]$

Explanation:

$f \left(x\right) = {x}^{3} + 5 {x}^{2} - 4 x - 20$

${x}^{3} + 5 {x}^{2} - 4 x - 20 \ge 0 \mathmr{and} {x}^{2} \left(x + 5\right) - 4 \left(x + 5\right) \ge 0$

or $= \left(x + 5\right) \left({x}^{2} - 4\right) \ge 0 \mathmr{and} \left(x + 5\right) \left(x + 2\right) \left(x - 2\right) \ge 0 \therefore$

$f \left(x\right) = \left(x + 5\right) \left(x + 2\right) \left(x - 2\right)$

Critical points are $x = - 5 , x = - 2 , x = 2$

$\therefore f \left(- 5\right) = f \left(- 2\right) = f \left(2\right) = 0$

Sign chart: When $x < - 5$ sign of $f \left(x\right)$ is (-)(-)(-)=(-) ; <0

When $- 5 < x < - 2$ sign of $f \left(x\right)$ is (+)(-)(-)=(+) ; >0

When $- 2 < x < 2$ sign of $f \left(x\right)$ is (+)(+)(-)=(-) ; <0

When $x > 2$ sign of $f \left(x\right)$ is (+)(+)(+)=(+) ; >0

Solution: $- 5 \le x \le - 2 \mathmr{and} x > 2$ . In interval notation:

$x | \left[- 5 , - 2\right] \cup \left[2 , \infty\right]$

graph{x^3+5x^2-4x-20 [-40, 40, -20, 20]} [Ans]