Answer
The answer is $\frac{2x^2+5x-2}{(x-5)(x+2)}$.
Work Step by Step
The given expression is
$\frac{2x+1}{x-5}-\frac{4}{x^2-3x-10}$
Factor the denominator in the second terms.
$=x^2-3x-10$
$=x^2-5x+2x-10$
$=x(x-5)+2(x-5)$
$(x-5)(x+2)$ Plug into the expression.
$=\frac{2x+1}{x-5}-\frac{4}{(x-5)(x+2)}$
The least common denominator for both terms is
$=(x-5)(x+2)$
Multiply and divide the first terms by $x+2$.
$=\frac{(2x+1)(x+2)}{(x-5)(x+2)}-\frac{4}{(x-5)(x+2)}$
Clear the parentheses in the first term.
$=\frac{2x^2+4x+x+2}{(x-5)(x+2)}-\frac{4}{(x-5)(x+2)}$
Both denominators are equal therefore, Subtract both numerators.
$=\frac{2x^2+4x+x+2-4}{(x-5)(x+2)}$
Simplify.
$=\frac{2x^2+5x-2}{(x-5)(x+2)}$.
