Another Inequality

Here , in this equation $\dfrac {(x-1)(x+4)(x+2)}{(x^2-9)(x+8)}<0$ = $\dfrac {(x-1)(x+4)(x+2)}{(x-3)(x+3)(x+8)}<0$

Multiplying $(x-3)(x+3)(x+8)$ in Numerator And Denominator, We get

$\dfrac {(x-1)(x+4)(x+2)(x-3)(x+3)(x+8)}{(x-3)^2(x+3)^2(x+8)^2}<0$

Hence As Denominator is $> 0$ ( As all are squares)

We get

$(x-1)(x+4)(x+2)(x-3)(x+3)(x+8)<0$

Hence,

By wavy Curve Method,

It looks something like this,

Now as we have to find for $< 0$ , We will take the negetive part

hence $x \epsilon$ (- 8 , -4) U (-3 , -2) U (1, 3))

Hence now see, x must be integers

So Between $-8$ and $-4$ , we get $-5, -6, -7$ ----> $\boxed{3}$ Solutions

For $-3$ to $-2$ , $\boxed{0}$ solutions as No integer is there in between

and Between $1$ to $3$ we have $2$ , So $\boxed{1}$ solutions

Hence Total Solutions = $3+ 1 + 0$ = $\boxed{4}$