How do you solve #(x-2)(x+1)(x-5)>=0# using a sign chart?

Question

2291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-25T11:57:42

Solution: # -1 <= x < 2 and x >=5 or [-1,2] uu [5,oo) #

#f(x)= (x-2) (x+1) (x-5)>= 0# . Crititical numbers

are # x=-1 , x=2 ,x=5#. Since at those numbers #f(x)=0#

Sign chart:

When #x< -1# sign of #(x-2)(x+1)(x-5) # is # (-) * (-)* (-) = (-) ; < 0#

When # -1 < x < 2 # sign of #(x-2)(x+1)(x-5) # is # (-) * (+)* (-) = (+) ; > 0#

When # 2 < x < 5 # sign of # (x-2)(x+1)(x-5) # is # (+) * (+)* (-) = (-) ; < 0#

When # x > 5 # sign of #(x-2)(x+1)(x-5) # is # (+) * (+)* (+) = (+) ; > 0#

Solution: # -1 <= x < 2 and x >=5 or [-1,2] uu [5,oo) #

graph{(x-2)(x+1)(x-5) [-40, 40, -20, 20]} [Ans]