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Partial fractions

Algebra
A-level Maths (9MA0)
ORIGINAL

One fraction per linear factor, then substitute the clever values of x to pick off each constant.

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Question

1.

Express 3x+1(x−1)(x+2)\dfrac{3x + 1}{(x - 1)(x + 2)} in partial fractions.

Every step worked, with the reasoning.

  1. 1
    3x+1(x−1)(x+2)≡Ax−1+Bx+2\dfrac{3x+1}{(x-1)(x+2)} \equiv \dfrac{A}{x-1} + \dfrac{B}{x+2}

    Set up one fraction per linear factor.

  2. 2
    3x+1≡A(x+2)+B(x−1)3x + 1 \equiv A(x+2) + B(x-1)

    Multiply both sides by (x−1)(x+2)(x-1)(x+2).

  3. 3
    x=1:  4=3A⇒A=43x = 1:\ \ 4 = 3A \Rightarrow A = \dfrac{4}{3}

    Substitute x=1x = 1 to kill the BB term.

  4. 4
    x=−2:  −5=−3B⇒B=53x = -2:\ \ -5 = -3B \Rightarrow B = \dfrac{5}{3}

    Substitute x=−2x = -2 to kill the AA term.

  5. 5
    3x+1(x−1)(x+2)≡43(x−1)+53(x+2)\dfrac{3x+1}{(x-1)(x+2)} \equiv \dfrac{4}{3(x-1)} + \dfrac{5}{3(x+2)}

    Write out the result.

Answer: 43(x−1)+53(x+2)\dfrac{4}{3(x-1)} + \dfrac{5}{3(x+2)}

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