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

1 Answer
Binayaka C.
Nov 25, 2017

Solution: -1 <= x < 2 and x >=5 or [-1,2] uu [5,oo) −1≤x<2andx≥5or[−1,2]∪[5,∞)

Explanation:

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

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

Sign chart:

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

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

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

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

Solution: -1 <= x < 2 and x >=5 or [-1,2] uu [5,oo) −1≤x<2andx≥5or[−1,2]∪[5,∞)

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

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