Q.Resolve $\dfrac{x+3}{(x-1)^2(x+2)}$ into partial fractions.

The factor $(x-1)$ is repeated, so it contributes two separate terms — one for $(x-1)$ and one for $(x-1)^2$ — in addition to the term for $(x+2)$.

Write $\dfrac{x+3}{(x-1)^2(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}$, giving $x+3 = A(x-1)(x+2)+B(x+2)+C(x-1)^2$ after clearing denominators.

Since $(x-1)$ is a repeated linear factor, write $\dfrac{x+3}{(x-1)^2(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}$. Multiplying both sides by $(x-1)^2(x+2)$:

Put $x=1$: $4 = B(3) \Rightarrow B=\dfrac{4}{3}$.

Put $x=-2$: $1 = C(9) \Rightarrow C=\dfrac{1}{9}$.

$A$ cannot be isolated by a root substitution alone (both roots used are exhausted), so compare the coefficient of $x^2$ on both sides. Expanding the right side, the $x^2$ terms come only from $A(x-1)(x+2)$ and $C(x-1)^2$, giving coefficient $A+C$. The left side has no $x^2$ term, so $0 = A+C \Rightarrow A = -C = -\dfrac{1}{9}$.

So $\dfrac{x+3}{(x-1)^2(x+2)} = -\dfrac{1}{9(x-1)} + \dfrac{4}{3(x-1)^2} + \dfrac{1}{9(x+2)}$.

Check (independent verification): substitute $x=0$. Original: $\dfrac{3}{(1)(2)} = \dfrac{3}{2}$. Decomposed: $-\dfrac{1}{9(-1)}+\dfrac{4}{3(1)}+\dfrac{1}{9(2)} = \dfrac{1}{9}+\dfrac{4}{3}+\dfrac{1}{18} = \dfrac{2}{18}+\dfrac{24}{18}+\dfrac{1}{18} = \dfrac{27}{18}=\dfrac{3}{2}$. Both sides agree.

The most frequent error is writing only $\dfrac{A}{x-1}+\dfrac{C}{x+2}$ and forgetting the extra $\dfrac{B}{(x-1)^2}$ term that a repeated factor requires. A second common error is trying to find all three unknowns purely by root substitution, when only two roots exist — the third unknown needs a coefficient comparison.