Since multiplying both sides of an inequality by a positive number keeps the sign unchanged, we can rewrite
{(x+1)(x+2)}/(x-3) > 0(x+1)(x+2)x−3>0
as
(x-3)^2 {(x+1)(x+2)}/(x-3) > (x-3)^2 times 0 = 0(x−3)2(x+1)(x+2)x−3>(x−3)2×0=0
That is
(x+1)(x+2)(x-3) >0 (x+1)(x+2)(x−3)>0
Now, the function f(x) = (x+1)(x+2)(x-3) f(x)=(x+1)(x+2)(x−3) has zeros at x=-2x=−2, x=-1x=−1 and x=+3x=+3. These are the points where the function can change sign. We are, of course, looking for regions where the function is positive.
When x<-2x<−2, all three factors are negative, and so the product is negative. At x=-2x=−2, the function is zero - in both cases, the inequality is not satisfied.
When -2 < x < -1 −2<x<−1, (x+2)(x+2) is positive, while the other two factors are negative - so our function f(x)f(x) is positive!
For -1 < x < 3−1<x<3, (x+1)(x+1) and (x+2)(x+2) are positive, but (x-3)(x−3) is negative - leading to f(x)<0f(x)<0
Finally, for x > 3x>3, all three factors are positive - and so f(x) > 0f(x)>0
Thus, the inequality is satisfied for both -2 <x < -1 −2<x<−1 and x > 3x>3
This can also be seen at a glance from the graph of f(x)f(x) below :
graph{(x+1)(x-2)(x+3) [-4, 4, -2, 6]}
