(x-1)(x+2)<0?

There are 2 critical values for this inequality: where x = -2 and where x = 1.

If we choose a test value that is less than -2, like - 5, we can see that the inequality is false.
(-5 - 1)(-5 + 2) < 0
(-6)(-3) < 0
18 < 0

If we choose a test value that is between -2 and 1, like 0, we can see that the inequality is true.
(0 - 1)(0 + 2) < 0
(-1)(2) < 0
-2 < 0

If we choose a test value that is larger than 1, like 5, we can see that the inequality is false.
(5 - 1)(5 + 2) < 0
(4)(7) < 0
28 < 0

So the only region that had a test value that made the equation true was between -2 and 1. Therefore, the answer to the inequality is -2 < x < 1.

(x - 1)(x + 2) < 0
x < 1 or -2

I'm going to assume this is Algebra or Analytic Geometry. If it is calculus, you will have quicker and more general methods of getting an answer.

y = (x-1)(x+2) is a parabola, as it has an x^2 term in it if you multiply it out. It crosses the x-axis at exactly two places. You can see the two places where y=0 just by looking at the equation.

Now, you must determine whether the places where y is less than 0 are between those two points, or outside of those two points. You could determine this by trying a few values of x.

p.s. Looks like Bryanna answered while I was typing. I like her answer better.

The two answers above are vague..let me try hold on let me type this up. nevermind the person above me got it.

(x - 1)(x + 2) < 0
x - 1 > 0 and x + 2 < 0 . . . or . . . x - 1 < 0 and x + 2 > 0
x > 1 and x < - 2 . . . or . . . x < 1 and x > - 2
. . . . . ø . . . . . . . . . . U . . . . . . (- 2 ; 1)

A = (- 2 ; 1)

Alejandra