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

Question

1910 views around the world

You can reuse this answer
Creative Commons License

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]