Given :
[tex]\frac{5x}{x^2-x-6}-\frac{4}{x^2+4x+4}[/tex]We need to factor the polynomial in the denominator.
[tex]\frac{5x}{x^2-x-6}-\frac{4}{x^2+4x+4}=\frac{5x}{x^2-3x+2x-6}-\frac{4}{x^2+2\times2x+2^2}[/tex][tex]=\frac{5x}{x(x^{}-3)+2(x-3)}-\frac{4}{(x+2)^2}[/tex][tex]=\frac{5x}{(x^{}-3)(x+2)}-\frac{4}{(x+2)^2}[/tex]The least common multiple of (x-3)(x+2) and (x+2)(x+2) is (x-3)(x+2)(x+2), so making the denominatore (x-3)(x+2)(x+2).
[tex]=\frac{5x(x+2)}{(x^{}-3)(x+2)^2}-\frac{4(x-3)}{(x-3)(x+2)^2}[/tex][tex]=\frac{5x^2+10x}{(x^{}-3)(x+2)^2}-\frac{4x-12}{(x-3)(x+2)^2}[/tex][tex]=\frac{5x^2+10x-4x+12}{(x^{}-3)(x+2)^2}[/tex][tex]=\frac{5x^2+6x+12}{(x^{}-3)(x+2)^2}[/tex]Hence the difference of the given is
[tex]\frac{5x^2+6x+12}{(x^{}-3)(x+2)^2}[/tex]