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Partial fractions · Lesson 1 of 2

Partial fractions with linear factors

Split a fraction whose denominator has two or three different linear factors, find where it is undefined, and divide first when the top is not of lower degree.

18 minYou should already know: Polynomials & quadratic roots Expressions, formulae & change of subject
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In General Maths you met partial fractions for the first time (see algebraic fractions): adding fractions, run backwards. You split 5x−1(x−1)(x+1)\dfrac{5x - 1}{(x - 1)(x + 1)} into 2x−1+3x+1\dfrac{2}{x - 1} + \dfrac{3}{x + 1} by making each bracket zero in turn.

5x − 1(x − 1)(x + 1)=Ax − 1+Bx + 1x = 1 gives A = (5 − 1) ÷ 2 = 2x = −1 gives B = (−5 − 1) ÷ (−2) = 3
Partial fractionsMake each bracket zero in turn to find A and B

Further Maths uses the same method, but the denominator often comes unfactorised, it may have three brackets, and the top may be “too big”. This lesson takes each of those in turn. Lesson 2 deals with brackets that repeat and quadratics that don’t factorise.

More: fractions over one denominator

Two linear factors

The method, one move at a time:

  1. Factorise the denominator, if it isn’t already.
  2. Write one fraction for each bracket: A…+B…\dfrac{A}{\ldots} + \dfrac{B}{\ldots}.
  3. Multiply through by the whole denominator. This gives an identity, true for every xx.
  4. Put in the value of xx that makes one bracket zero. Every other term vanishes, leaving one constant.
  5. Repeat for each bracket, then check with a value you haven’t used, such as x=0x = 0.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q11 (b)

A function gg is defined by g(x)=3−4xx2+x−6g(x) = \dfrac{3 - 4x}{x^2 + x - 6}. Express g(x)g(x) in partial fractions.

  1. Factorise the bottom

    • The numbers are 3 and −2-2.
    • So x2+x−6=(x−2)(x+3){x^2 + x - 6 = (x - 2)(x + 3)}.

    Think first. Two numbers that multiply to −6 and add to 1?

  2. Set up and multiply through

    • Write 3−4x(x−2)(x+3)=Ax−2+Bx+3{\frac{3 - 4x}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}}.
    • Multiply through by (x−2)(x+3)(x - 2)(x + 3): 3−4x=A(x+3)+B(x−2){3 - 4x = A(x + 3) + B(x - 2)}.

  3. Make each bracket zero

    • Put x=2x = 2: 3−8=A(5)+0{3 - 8 = A(5) + 0}.
    • So 5A=−5{5A = -5}, and A=−1{A = -1}.
    • Put x=−3x = -3: 3+12=0+B(−5){3 + 12 = 0 + B(-5)}.
    • So −5B=15{-5B = 15}, and B=−3{B = -3}.

    Think first. Which x makes the B term vanish? Which makes the A term vanish?

  4. Answer and check

    • g(x)=−1x−2−3x+3{g(x) = -\frac{1}{x - 2} - \frac{3}{x + 3}}.
    • At x=0x = 0 the fraction is 3−6=−12{\frac{3}{-6} = -\frac12}.
    • The answer gives −1−2−33=12−1=−12{-\frac{1}{-2} - \frac33 = \frac12 - 1 = -\frac12} ✓.

    Think first. Check with x = 0: do both sides agree?

Pick a value of xx and watch which terms vanish:

Make one bracket zeroPick a value of x

4x + 5 = A(x − 1) + B(x + 2)

?A?B
Multiply through by the whole denominator. The identity is true for every x, so choose the x that makes one bracket zero.

More: two linear factors

When a bracket is 2x − 1 or 3x + 1

The value that makes 2x−12x - 1 zero is x=12x = \frac12, not x=1x = 1. Solve “bracket =0= 0” each time, and expect fractions in the working.

More: brackets such as 3x + 1

Three factors, and where the fraction is undefined

A fraction is undefined where its denominator is zero, so those values of xx are exactly the ones that make a bracket zero. Factorise fully first: x2−1x^2 - 1 is two brackets, (x−1)(x+1)(x - 1)(x + 1).

With three brackets, write three fractions. When you multiply through, each constant is multiplied by the other two brackets.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q10 (a)

If f(x)=2x−3(x2−1)(x+2)f(x) = \dfrac{2x - 3}{(x^2 - 1)(x + 2)}, (i) find the values of xx for which f(x)f(x) is undefined; (ii) express f(x)f(x) in partial fractions.

  1. Where it is undefined

    • Factorise: (x2−1)(x+2)=(x−1)(x+1)(x+2){(x^2 - 1)(x + 2) = (x - 1)(x + 1)(x + 2)}.
    • The bottom is zero when x=1{x = 1}, x=−1{x = -1} or x=−2{x = -2}.
    • So f(x)f(x) is undefined at x=1,−1,−2x = 1, -1, -2.

    Think first. Factorise x² − 1. Which x values make the bottom zero?

  2. Set up

    • Write Ax−1+Bx+1+Cx+2{\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2}}.
    • Multiply through by the whole denominator. Each constant is multiplied by the other two brackets:
    2x−3=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1)\begin{aligned} 2x - 3 = {} & A(x + 1)(x + 2) \\ & + B(x - 1)(x + 2) \\ & + C(x - 1)(x + 1) \end{aligned}

  3. Make each bracket zero

    • x=1x = 1: −1=A(2)(3)=6A{-1 = A(2)(3) = 6A}, so A=−16{A = -\frac16}.
    • x=−1x = -1: −5=B(−2)(1)=−2B{-5 = B(-2)(1) = -2B}, so B=52{B = \frac52}.
    • x=−2x = -2: −7=C(−3)(−1)=3C{-7 = C(-3)(-1) = 3C}, so C=−73{C = -\frac73}.

    Think first. Put x = 1, then x = −1, then x = −2.

  4. Answer

    f(x)=−16(x−1)+52(x+1)−73(x+2)\begin{aligned} f(x) = {} & -\frac{1}{6(x - 1)} + \frac{5}{2(x + 1)} \\ & - \frac{7}{3(x + 2)} \end{aligned}

More: three factors

When the top is too big: divide first

Partial fractions only work when the top has a lower degree than the bottom (a “proper” fraction). If the degree of the top is the same or higher, divide first, just as 175=325\frac{17}{5} = 3\frac25:

175=3+25whole part + remainderx² + 3x + 5x + 1=x + 2+3x + 1
Divide firstWhole part, plus a proper fraction to split

Then split the remainder fraction as usual.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q10 (a)

Resolve x3−x2−4x2−1\dfrac{x^3 - x^2 - 4}{x^2 - 1} into partial fractions.

  1. Is it proper?

    • The top has degree 3 and the bottom degree 2.
    • So divide first.

    Think first. Degree of the top? Of the bottom?

  2. Divide

    • xx times the bottom: x(x2−1)=x3−x{x(x^2 - 1) = x^3 - x}.
    • Take it away: (x3−x2−4)−(x3−x)=−x2+x−4{(x^3 - x^2 - 4) - (x^3 - x) = -x^2 + x - 4}.
    • −1-1 times the bottom: −1(x2−1)=−x2+1{-1(x^2 - 1) = -x^2 + 1}.
    • Take it away: (−x2+x−4)−(−x2+1)=x−5{(-x^2 + x - 4) - (-x^2 + 1) = x - 5}.
    • So x3−x2−4x2−1=x−1+x−5x2−1{\frac{x^3 - x^2 - 4}{x^2 - 1} = x - 1 + \frac{x - 5}{x^2 - 1}}.

    Think first. What times x² gives x³?

  3. Split the remainder

    • Write x−5(x+1)(x−1)=Ax+1+Bx−1{\frac{x - 5}{(x + 1)(x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1}}, so x−5=A(x−1)+B(x+1){x - 5 = A(x - 1) + B(x + 1)}.
    • x=1x = 1: −4=2B{-4 = 2B}, so B=−2{B = -2}.
    • x=−1x = -1: −6=−2A{-6 = -2A}, so A=3{A = 3}.

    Think first. x² − 1 = (x + 1)(x − 1). Put x = 1, then x = −1.

  4. Answer

    x3−x2−4x2−1=x−1+3x+1−2x−1\frac{x^3 - x^2 - 4}{x^2 - 1} = x - 1 + \frac{3}{x + 1} - \frac{2}{x - 1}

More: divide first

Your turn

WAEC 2011 · Paper 2 · Q12 (b)

  1. (b)

    Express 3x+2x2+x−2\dfrac{3x + 2}{x^2 + x - 2} in partial fractions.

Worked solution (try it first)

(b)

  1. Factorise the bottom: x2+x−2=(x+2)(x−1)x^2 + x - 2 = (x + 2)(x - 1).
  2. Write Ax+2+Bx−1\dfrac{A}{x + 2} + \dfrac{B}{x - 1} and multiply through: 3x+2=A(x−1)+B(x+2)3x + 2 = A(x - 1) + B(x + 2).
  3. Put x=1x = 1: 5=3B5 = 3B, so B=53B = \frac53.
  4. Put x=−2x = -2: −4=−3A-4 = -3A, so A=43A = \frac43.
  5. So the answer is 43(x+2)+53(x−1)\dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)}.

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