Q.Resolve $\dfrac{3x+5}{(x-1)(x+2)}$ into partial fractions.

This is a proper fraction whose denominator has two distinct, non-repeated linear factors, so it splits into two simple fractions with constants found by substitution.

Write $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}$. Multiplying both sides by $(x-1)(x+2)$ gives $3x+5 = A(x+2)+B(x-1)$. Putting $x=1$: $8=3A \Rightarrow A=\dfrac{8}{3}$. Putting $x=-2$: $-1=-3B \Rightarrow B=\dfrac{1}{3}$.

The denominator $(x-1)(x+2)$ has two distinct, non-repeated linear factors, so write $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}$ for constants $A,B$ to be found.

Multiplying both sides by $(x-1)(x+2)$: $3x+5 = A(x+2)+B(x-1)$.

Put $x=1$ (this zeroes out the $B$ term): $3(1)+5 = A(1+2)+0 \Rightarrow 8 = 3A \Rightarrow A=\dfrac{8}{3}$.

Put $x=-2$ (this zeroes out the $A$ term): $3(-2)+5 = 0+B(-2-1) \Rightarrow -1=-3B \Rightarrow B=\dfrac{1}{3}$.

So $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{8}{3(x-1)} + \dfrac{1}{3(x+2)}$.

Check (independent verification): substitute $x=0$ into both the original expression and the decomposed answer. Original: $\dfrac{0+5}{(-1)(2)} = -\dfrac{5}{2}$. Decomposed: $\dfrac{8}{3(-1)} + \dfrac{1}{3(2)} = -\dfrac{8}{3}+\dfrac{1}{6} = -\dfrac{16}{6}+\dfrac{1}{6} = -\dfrac{15}{6} = -\dfrac{5}{2}$. Both sides agree, confirming $A$ and $B$ are correct.

Alternatively, equate coefficients after expanding $3x+5=A(x+2)+B(x-1) = (A+B)x + (2A-B)$: this gives $A+B=3$ and $2A-B=5$. Adding the two equations, $3A=8 \Rightarrow A=\dfrac{8}{3}$, then $B=3-\dfrac{8}{3}=\dfrac{1}{3}$ — the same values as the substitution method, confirming the result a second way.

A common slip is forgetting to multiply both sides by the full denominator before substituting values of $x$, or choosing a value of $x$ that does not zero out one of the unknowns (which fails to isolate a single constant in one step). Also watch the sign carefully when substituting $x=-2$.