Find all the values of x that satisfy the inequality below.

$$\dfrac{\left(x\ -\ 3\right)^2\left(x\ ^2\ +\ 9x\ +\ 20\right)}{\left(x^2\ +\ 2x\ -\ 15\right)}\ \le\ 0$$

Given inequality is, $$\dfrac{\left(x\ -\ 3\right)^2\left(x\ ^2\ +\ 9x\ +\ 20\right)}{\left(x^2\ +\ 2x\ -\ 15\right)}\ \le\ 0$$

$$x\ ^2\ +\ 9x\ +\ 20\ $$ can be written as,

$$x\ ^2\ +\ 4x\ \ +\ 5x\ +\ 20\ $$  =  $$x\ \left(x\ +\ 4\right)\ \ +\ 5\left(x\ +\ 4\right)\ $$  =  $$\left(x\ +\ 5\right)\left(x\ +\ 4\right)\ $$

$$x\ ^2\ +\ 2x\ -\ 15\ $$ can be written as,

$$x\ ^2\ -\ 3x\ \ +\ 5x\ -\ 15\ $$ = $$x\ \left(x\ -\ 3\right)\ \ +\ 5\left(x\ -\ 3\right)\ $$ = $$\left(x\ +\ 5\right)\left(x\ -\ 3\right)\ $$

The inequality can be written as,

$$\dfrac{\left(x\ -\ 3\right)^2\left(x\ +\ 4\right)\left(x\ +\ 5\right)}{\left(x-\ 3\right)\left(x\ +\ 5\right)}\ \le\ 0$$

We can cancel the terms (x - 3) and (x + 5) from the numerator and denominator with the condition that x cannot take the values 3 and - 5.

After cancellation, the equation becomes,

$$\left(x\ -\ 3\right)\left(x\ +\ 4\right)\ \le\ 0$$

So, the values x can take are [-4, 3). We cannot include 3 as we already obtained the condition above.

So, the correct answer is option C.