Answer
$\frac{(x-1)}{(x+2)}; x \ne -1,-2;$
Work Step by Step
$\frac{4x^{2}+x-6}{x^{2}+3x+2} - \frac{3x}{x+1} + \frac{5}{x+2}$
Factors of $x^{2}+3x+2$ are $(x+1)(x+2)$. The given expression becomes
$=\frac{4x^{2}+x-6}{(x+1)(x+2)} - \frac{3x}{x+1} + \frac{5}{x+2}; x \ne -1,-2;$
Taking LCD,
$=\frac{4x^{2}+x-6- 3x(x+2)+5(x+1)}{(x+1)(x+2)}; x \ne -1,-2;$
$=\frac{4x^{2}+x-6- 3x^{2}-6x+5x+5}{(x+1)(x+2)}; x \ne -1,-2;$
Adding like terms we get
$=\frac{(4-3)x^{2}+(1-6+5)x+(-6+5)}{(x+1)(x+2)}; x \ne -1,-2;$
$=\frac{x^{2}-1}{(x+1)(x+2)}; x \ne -1,-2;$
$(x^{2}- 1) $ is the difference of squares, the factors are $(x+1)(x-1)$
$[a^{2}-b^{2} = (a+b)(a-b)]$
$=\frac{(x+1)(x-1)}{(x+1)(x+2)}; x \ne -1,-2;$
Divide out common factors.
$= \frac{(x-1)}{(x+2)}; x \ne -1,-2;$
