WAEC Further Mathematics 2023 Paper

Concept:This is a partial fractions problem where the numerator is compared after rewriting the right-hand side with a common denominator.Explanation:Start with the given identity:$\frac{3x+4}{(x-2)(x+3)} \equiv \frac{P}{x+3}+\frac{Q}{x-2}$Combine the fractions on the right:$\frac{P}{x+3}+\frac{Q}{x-2}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}$Since the denominators are equal, compare the numerators:$3x+4=P(x-2)+Q(x+3)$Expand the right-hand side:$3x+4=Px-2P+Qx+3Q$Group like terms:$3x+4=(P+Q)x+(-2P+3Q)$Now equate the coefficients of $x$ and the constant term:$P+Q=3$$-2P+3Q=4$From the first equation:$P=3-Q$Substitute $P=3-Q$ into $-2P+3Q=4$:$-2(3-Q)+3Q=4$$-6+2Q+3Q=4$$-6+5Q=4$$5Q=10$$Q=2$Answer:$Q=2$Therefore, the correct option is A. 2.