(2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)
If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?
And why can't X = -3\2?
$\displaystyle \frac{2x^2 - 7x - 15}{2x^3 + 13x^2 + 15x}$
$\displaystyle = \frac{2x^2 - 10x + 3x - 15}{x(2x^2 + 13x + 15)}$
$\displaystyle = \frac{2x(x - 5) + 3(x - 5)}{x(2x^2 + 10x + 3x + 15)}$
$\displaystyle = \frac{(x - 5)(2x + 3)}{x[2x(x + 5) + 3(x + 5)]}$
$\displaystyle = \frac{(x - 5)(2x + 3)}{x(x + 5)(2x + 3)}$
Can you see here that you can't let $\displaystyle x = -\frac{3}{2}$? If you did the denominator would be $\displaystyle 0$.
Now cancelling gives
$\displaystyle = \frac{x - 5}{x(x + 5)}$
$\displaystyle = \frac{x - 5}{x^2 + 5x}$.
$\displaystyle \frac{2x^2-7x-15}{2x^3+13x^2+15x}=\frac{(2x+3)(x-5)}{x(2x+3)(x+5)}$
This has no value if x=0, x= -5 and 2x = -3.
You can cancel the term (2x+3) only if (2x+3) is not zero.
Since in that case you have 0/0.
However, if x>0, you can cancel those terms and then we can say it's the same as
$\displaystyle \frac{x-5}{x(x+5)}$