Please,solve these two math problems for me...????

Question

1-) (x+3) (x-2)<0

2-)8^2/3=

Answer 1

1. (x+3)(x-2) < 0
Convert to equality:
(x+3)(x-2) = 0
set (x+3) and (x-2) each to be equal to 0 and solve:
x + 3 = 0 x = -3
x - 2 = 0 x = 2
now look at where x must be by multipling the 2 binomials:
x^2 + x - 6 < 0 and try values on both sides and in the middle of your range:
x = 3
3^2 + 3 - 6 = 6 < 0 is not true, so that won't work.
x = 0
0^2 + 0 - 6 = -6 < 0 is true, so it works for values less than 2.
x = -4
(-4)^2 + (-4) - 6 = 6 < 0 is not true, so values less that -3 won't work.
This means that your domain where this is true is:
-3 > x > 2

2.
8^(2/3) = (8^(1/3))^2 = (2)^2 = 4.

Answer 2

1> It has 2 solutions

i> x + 3 < 0
=> x < -3 => X varies from -4 to - infinity
ii> x-2 < 0
=> x < 2
=> X can vary from any value less than 2 till negative infinity

2> 8^2/3 = 2^(3*2/3)
=> 8^2/3 = 2^ 2 = 4

Answer 3

1) For a product to be less than 0, one number should be negative while the other is positive. It's not possible for x+3 to be negative and x-2 positive for the same x, so we want x+3 to be positive and x-2 negative. For x+3 to be positive, x must be greater than -3. For x-2 to be negative, x must be less than 2. So the solution is -3 < x < 2.

2) There's two ways to do this. Either take the cube root of 8 (2) and square it to get 4, or square 8 to get 64, then take the cube root of 64 which is 4. Either way, the answer is 4.

Answer 4

1) the graph of y = (x+3)(x-2) opens upward, has x-intercepts at -3 and 2, and so is below the x axis, y < 0, for
-3 < x < 2.

2) 8^(2/3) = [8^(1/3)]² = 2² = 4

Answer 5

u divide the x -line in three parts or intervals they are x<-3; x>2 and -3

Answer 6

I know 1-)
(x+3) (x-2) <0
x squared + 3x -2x -6 <0
x squared + x -6 <0

I don't really remember if the < stays pointing the same way though. I think it does, but I'm not sure, so don't hold me to it.