Simplify the numerator of the fraction.

The expression becomes $2 - \left(\frac{y}{0}\right)$.

Division by zero is undefined. Therefore, the entire expression is undefined.

• To find $A$, set $x=1$: $$1 = A(1+2) + B(1-1)$$ $$1 = 3A + 0$$ $$A = \frac{1}{3}$$
• To find $B$, set $x=-2$: $$1 = A(-2+2) + B(-2-1)$$ $$1 = 0 + B(-3)$$ $$B = -\frac{1}{3}$$

For a repeated linear factor $(x-2)^2$, the decomposition includes terms $\frac{A}{x-2}$ and $\frac{B}{(x-2)^2}$.

• To find $B$, set $x=2$: $$3 = A(2-2)(2+1) + B(2+1) + C(2-2)^2$$ $$3 = A(0) + B(3) + C(0)$$ $$3 = 3B \Rightarrow B = 1$$
• To find $C$, set $x=-1$: $$3 = A(-1-2)(-1+1) + B(-1+1) + C(-1-2)^2$$ $$3 = A(0) + B(0) + C(-3)^2$$ $$3 = 9C \Rightarrow C = \frac{3}{9} = \frac{1}{3}$$
• To find $A$, choose a convenient value for $x$, such as $x=0$, and substitute the known values of $B$ and $C$: $$3 = A(0-2)(0+1) + B(0+1) + C(0-2)^2$$ $$3 = A(-2)(1) + B(1) + C(-2)^2$$ $$3 = -2A + B + 4C$$ Substitute $B=1$ and $C=\frac{1}{3}$: $$3 = -2A + 1 + 4\left(\frac{1}{3}\right)$$ $$3 = -2A + 1 + \frac{4}{3}$$ $$3 - 1 - \frac{4}{3} = -2A$$ $$2 - \frac{4}{3} = -2A$$ $$\frac{6}{3} - \frac{4}{3} = -2A$$ $$\frac{2}{3} = -2A$$ $$A = \frac{2}{3} \times \left(-\frac{1}{2}\right) = -\frac{1}{3}$$