## anonymous 5 years ago Solve the inequality, (x+3)(x-2)>0. Express your answer using set notation.

1. anonymous

Imagine x is a big negative number? What will the result of this expression be?

2. myininaya

(x+3)(x-2) is 0 when x=-3 and 2. So use test points before and after -3 and 2 to see what are happening to the factors. so plug in some number before -3 like -4 so -4+3=-1 -4-2=-6 (-1)(-6)=6>0 so part of the solution is x is (-inf,-3) now plug in a number between -3 and 2 like 0 so 0+3=3 0-2=-2 (3)(-2)=-6<0 so the interval (-3,2) is not apart of the solution now you do the last interval

3. anonymous

If x is a big negative numer x+3 will still be big and negative, and x-2 will also be negative. A negative times a negative is positive. So big negative values for x are ok.

4. anonymous

Now as x gets close to -3 from the negative infinity side, the (x+3) factor will start getting closer and closer to 0. Once x = -3 we see that the expression becomes 0, so this is not allowed.

5. anonymous

So our first interval is $$(-\infty,-3)$$. Now if x is between -3 and 2 what will this expression be?

6. anonymous

so the last interval would be, (2,inf)?

7. myininaya

is that part of the solution? let's check: 3+3=6 3-1=2 6(2)>0 so yes thats right so your whole answer is (-inf,3)U(2,inf)

8. myininaya

that 3 is suppose to be negative

9. myininaya

(-inf,-3)U(2,inf)

10. anonymous

is that part of the solution? let's check: 3+3=6 3-1=2*** 6(2)>0 so yes thats right so your whole answer is (-inf,3)U(2,in *** the 3-1, isnt it suppose to be 3-2?

11. myininaya

yes but we still get something positive no biggie little type-0