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

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Answer 1 · 2017-10-17T12:43:29

Solution: $-5 <= x <= -2 and x > 2$ . In interval notation:
$x| [-5,-2]uu [2,oo]$

$f(x)=x^3+5x^2-4x-20$

$x^3+5x^2-4x-20 >= 0 or x^2(x+5) -4(x+5) >=0$

or $=(x+5) (x^2-4) >=0 or(x+5)(x+2)(x-2) >=0 :.$

$f(x)=(x+5)(x+2)(x-2)$

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

$:.f(-5)=f(-2)=f(2)=0$

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

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

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

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

Solution: $-5 <= x <= -2 and x > 2$ . In interval notation:

$x| [-5,-2]uu [2,oo]$

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

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