(x+3)(x-5)<0, how to find the exact inequality of x?

Question

answer if you are very confident. tq

Answer 1

Check where one factor is negative and the other is positive

If x<-3

x+3 < -3 + 3 = 0
x-5 < -3 - 5 = -8 < 0
Both factors are negative, thus their product is positive

Likewise, when x>5
(x + 3)(x - 5) > 0

If x=-3 or x=5 (x + 3)(x - 5) = 0

Let's check -3 < x < 5

x + 3 < -3 + x = 0
x - 5 > 5 - 5 = 0

Because one of the factors is negative and the other is positive:
(x + 3)(x - 5) < 0

Answer 2

Well, first, if it were equal to 0, x=-3 and x=5.
Now we have to look at values x<-3, -35.
First x<-3: try -4.
(-4+3)(-4-5)=-1*-9=9 which is not <0.

SO, try -3 (0+3)(0-5)=3*-5=-15 which is <0.

Now try x>5, try 6
(6+3)(6-5)=9*1=9 which is not <0.

SO, the final answer is:
-3

Answer 3

To satisfy inequation must have:-
x + 3 > 0
x > - 3
AND
x - 5 < 0
x < 5

x : - 3 < x < 5 is solution.

Answer 4

x^2 -2x - 15 <0
(x+3) < 0
x < -3

(x-5) < 0
x < 5