Q.Resolve $\dfrac{x+7}{(x+1)(x-3)}$ into partial fractions.

This proper fraction has two distinct linear factors in the denominator, so the standard substitution method applies directly.

Write $\dfrac{x+7}{(x+1)(x-3)}=\dfrac{A}{x+1}+\dfrac{B}{x-3}$, giving $x+7=A(x-3)+B(x+1)$.

Write $\dfrac{x+7}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}$. Multiplying both sides by $(x+1)(x-3)$: $x+7=A(x-3)+B(x+1)$.

Put $x=-1$: $6 = A(-4) \Rightarrow A=-\dfrac{3}{2}$.

Put $x=3$: $10 = B(4) \Rightarrow B=\dfrac{5}{2}$.

So $\dfrac{x+7}{(x+1)(x-3)} = -\dfrac{3}{2(x+1)} + \dfrac{5}{2(x-3)}$.

Check (independent verification): substitute $x=0$. Original: $\dfrac{7}{(1)(-3)} = -\dfrac{7}{3}$. Decomposed: $-\dfrac{3}{2(1)}+\dfrac{5}{2(-3)} = -\dfrac{3}{2}-\dfrac{5}{6} = -\dfrac{9}{6}-\dfrac{5}{6}=-\dfrac{14}{6}=-\dfrac{7}{3}$. Both sides match, confirming the constants.

Sign errors when substituting a negative root, e.g. computing $A(x-3)$ at $x=-1$ as $A(-1-3)=-4A$ but then mis-simplifying $6=-4A$ to a wrong value of $A$.