Can anyone spot my mistake?
$$\frac{1-\frac{1}{2}z^{-1}}{1+\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}}$$ Set $x = z^{-1}$ $$\frac{1-\frac{1}{2}x}{1+\frac{3}{4}x+\frac{1}{8}x^{2}}$$ Multiply by 8/8 and factorise denominator $$\frac{8-4x}{(x+2)(x+4)}$$ $$\frac{8-4x}{(x+2)(x+4)}=\frac{A}{x+2}+\frac{B}{x+4}$$ $$8-4x = A(x+4) + B(x+2)$$ Set $x = -2$ $$A = 0$$ Set $x = -4$ $$B = 4$$ Giving $$\frac{4}{4+z^{-1}}$$
The solution says that the answer is
$$\frac{-3}{1+\frac{1}{4}z^{-1}} + \frac{4}{1+\frac{1}{2}z^{-1}}$$
You have made a mistake in finding your $ B$
The value for $B$ should be negative.