X(x+3)(x-5)>0

[tushar.ju43]

x(x+3)(x-5)>0

Which of the following inequalities specify values ​​of x that satisfy the inequality above?

Indicate all such inequalities.

A. x < -3
B. -3 < x < 0
C. 0 < x < 5
D. x > 5

[mubinmodi]

B & C because x cannot equal to -3 , 0 or 5, because on substitution answer will be zero

[mdaminurrahmansarker]

1. x > 0 or
2. x+3 > 0
   => x > -3 or
3. x-5 > 0
   => x > 5
   so answer would be B, and D, respectively.

[C.Koushik]

If you get an equation like a.b.c.d > 0 (or a.b.c.d < 0), all you gotta do is draw a number line. Mark all the points where the terms a, b, c, and d are zero. And mark the alternative regions + and -, and pick the regions you want.

For your question, the roots would be x=0,-3 and 5

1. Draw a number line and plot the zeroes like this
   
   
2. Select a point in any region, and find what the value of the expression will be.
   
   
   
   For instance, I chose 8 here. So The value of the expression will be 8(11)(3) = +264. So any number in the region (5, \infty) will give us a positive value for the expression. So mark the region as +ve.
3. Mark the other regions alternately with + and -
   
   
4. Now, from the question, you want regions where the value is +ve, so your answer is (-3,0) and (5,\infty)

NOTE: One exception for this method is when you have the same term repeated an even number of times, say x(x+3)^{2}(x+5). In that case, the equality does not switch to the opposite sign about the point -3

[tushar.ju43]

In option B, X is also less than 0, Does it make sense?

[mdaminurrahmansarker]

Yes, I guess. as the results of different values need to be positive [x(x+3)(x-5)>0].
in option B, it is clearly indicating that in that range no matter what number we can choose result would be positive. here is what I mean:
let, x = -2, then (-2) (-2+3) (-2-5) = (-2)(1)(-7) = 14 > 0.