How do you solve (x-2)(x+5) >0?

1 Answer
May 17, 2015

The signs of the two factors change at x=-5 and x=2. So the sign of their product (x-2)(x+5) will change at those two points.

In the case x < -5:

(x - 2) < 0 and (x + 5) < 0 so (x-2)(x+5) > 0

In the case x = -5:

(x-2)(x+5) = 0

In the case -5 < x < 2:

(x - 2) < 0 but (x+5) > 0 so (x-2)(x+5) < 0

In the case x = 2:

(x-2)(x+5) = 0

In the case x > 2:

(x-2) > 0 and (x+5) > 0 so (x-2)(x+5) > 0

Putting these cases together (x-2)(x+5) > 0 when x < -5 or x > 2.