Answer
$$A=1$$ $$B=-1$$ $$C=0$$
Work Step by Step
$\frac{2x+4}{(x-2)(x^2+4)}$ = $\frac{A}{x-2}$+$\frac{Bx+C}{x^2+4}$
Multiplying $(x-2)(x^2+4)$ on both sides:
$2x+4=A(x^2+4)+(Bx+C)(x-2)$
If we let $x=2$ on both sides, we can eliminate constants B and C to find constant A:
$4+4=A(4+4)+(Bx+C)(2-2)$
Thus, $8=8A$ and so $A=1$.
Collecting all the constant terms on both sides:
$4=4A-2C$
Substituting $A=1$, we find that $C=0$.
Similarly, collecting all the $x$ terms on both sides:
$2x=-2Bx+Cx$
If we compare only the coefficients of the $x$ terms, we arrive at the equation:
$2=-2B+C$
Substituting $C=0$, we find that $B=-1$.
