Partial Fractions

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These are the solutions to the WASSCE Further Mathematics past questions on Partial Fractions.
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Forms of Whole Fractions

(1.) Proper Fractions: Non-Repeated Linear Factors at the Denominator
(2.) Proper Fractions: Non-Repeated Non-Linear Factors at the Denominator
(3.) Proper Fractions: Repeated Linear Factors at the Denominator
(4.) Proper Fractions: Repeated Non-Linear Factors at the Denominator
(5.) Proper Fractions: Repeated/Non-repeated Linear/Non-Linear Factors at the Denominator
(6.) Improper Fractions

(2.) Express $\dfrac{9x}{(2x + 1)(x^2 + 1)}$ in partial fractions.

$ \dfrac{9x}{(2x + 1)(x^2 + 1)} \\[5ex] \text{Form: Non-repeated Linear and Non-linear Factor at the denominator} \\[3ex] = \dfrac{A}{2x + 1} + \dfrac{Bx + C}{x^2 + 1} \\[5ex] \dfrac{9x}{(2x + 1)(x^2 + 1)} = \dfrac{A(x^2 + 1) + (Bx + C)(2x + 1)}{(2x + 1)(x^2 + 1)} \\[5ex] \text{Same denominator, equate the numerators} \\[3ex] A(x^2 + 1) + (Bx + C)(2x + 1) = 9x \\[3ex] Ax^2 + A + 2Bx^2 + Bx + 2Cx + C = 9x \\[3ex] Ax^2 + 2Bx^2 + Bx + 2Cx + A + C = 9x \\[3ex] x^2(A + 2B) + x(B + 2C) + A + C = 0x^2 + 9x + 0 \\[3ex] A + 2B = 0 ...eqn.(1) \\[3ex] B + 2C = 9 ...eqn.(2) \\[3ex] A + C = 0 ...eqn.(3) \\[3ex] \text{From eqn.(3)} \\[3ex] C = -A ...eqn.(4) \\[3ex] \text{Substitute for C in eqn.(2)} \\[3ex] B + 2(-A) = 9 \\[3ex] B - 2A = 9 \\[3ex] -2A + B = 9 ...eqn.(5) \\[3ex] 2 * eqn.(1) \rightarrow 2A + 4B = 0 ...eqn.(6) \\[3ex] eqn.(5) + eqn.(6) \rightarrow 5B = 9 \\[3ex] B = \dfrac{9}{5} \\[5ex] \text{Substitute for B in eqn.(1)} \\[3ex] A + 2\left(\dfrac{9}{5}\right) = 0 \\[5ex] A + \dfrac{18}{5} = 0 \\[5ex] A = -\dfrac{18}{5} \\[5ex] \text{Substitute for A in eqn.(4)} \\[3ex] C = -1 * -\dfrac{18}{5} \\[5ex] C = \dfrac{18}{5} \\[5ex] \implies \\[3ex] \dfrac{9x}{(2x + 1)(x^2 + 1)} \\[5ex] = \dfrac{-18}{5(2x + 1)} + \dfrac{9x + 18}{5(x^2 + 1)} \\[5ex] = -\dfrac{18}{5(2x + 1)} + \dfrac{9(x + 2)}{5(x^2 + 1)} $

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