Answer
a) $\frac{x^{2}+4x+12}{(x+2)(x^{2}+4)}$ = $\frac{1}{x+2}+\frac{4}{x^{2}+4}$
b) $\frac{2x^{2}+8x+24}{(x+2)^{2}(x^{2}+4)}$ = $\frac{1}{x+2}+\frac{2}{(x+2)^{2}}+\frac{-x+2}{x^{2}+4}$
c) $\frac{x^{2}-4x+8}{(x-1)^{2}(x-2)^{2}}$ = $\frac{-8}{x-2}+\frac{4}{(x-2)^{2}}+\frac{8}{x-1}+\frac{5}{(x-1)^{2}}$
d) $\frac{x^{4}-4x+8}{(x+2)(x^{2}+4)}$ = $x-2+\frac{4}{x+2}-\frac{4x-4}{x^{2}+4}$
Work Step by Step
a) $\frac{x^{2}+4x+12}{(x+2)(x^{2}+4)}$ = $\frac{1}{x+2}+\frac{4}{x^{2}+4}$
b) $\frac{2x^{2}+8x+24}{(x+2)^{2}(x^{2}+4)}$ = $\frac{1}{x+2}+\frac{2}{(x+2)^{2}}+\frac{-x+2}{x^{2}+4}$
c) $\frac{x^{2}-4x+8}{(x-1)^{2}(x-2)^{2}}$ = $\frac{-8}{x-2}+\frac{4}{(x-2)^{2}}+\frac{8}{x-1}+\frac{5}{(x-1)^{2}}$
d) $\frac{x^{4}-4x+8}{(x+2)(x^{2}+4)}$ = $x-2+\frac{4}{x+2}-\frac{4x-4}{x^{2}+4}$
