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# A-Level Maths | Partial Fractions
- URL: https://www.a-level-maths-tutor.com/a-level-maths-partial-fractions/
- Published: 2026-04-13T15:23:07.000Z
- Updated: 2026-07-11T13:39:28.000Z
- Author: Jaisul Naik

For distinct linear factors: $\\displaystyle\\frac{6x}{(x-1)(x+2)} = \\frac{A}{x-1} + \\frac{B}{x+2}$

Multiply both sides by $(x-1)(x+2)$, then substitute $x=1$ and $x=-2$ to solve for $A$ and $B$.

**Worked Example**

$\\displaystyle\\frac{6x}{(x-1)(x+2)} = \\frac{A}{x-1} + \\frac{B}{x+2}$

$6x = A(x+2) + B(x-1)$

$x=1: \\;\\; 6 = 3A \\;\\Rightarrow\\; A = 2$

$x=-2: \\;\\; -12 = -3B \\;\\Rightarrow\\; B = 4$

$\\displaystyle\\frac{6x}{(x-1)(x+2)} = \\frac{2}{x-1} + \\frac{4}{x+2}$

## Question 1

Express each of the following as partial fractions.

**a)** $\\displaystyle\\frac{6x}{(x-1)(x+2)}$

**b)** $\\displaystyle\\frac{7y-11}{(y+2)(y-3)}$

**c)** $\\displaystyle\\frac{19-4t}{(t+4)(t-3)}$

**d)** $\\displaystyle\\frac{w-22}{(w+2)(w-6)}$

**e)** $\\displaystyle\\frac{8z+7}{(z+2)(z-7)}$

Show Answers 

**a)** $\\displaystyle\\frac{2}{x-1} + \\frac{4}{x+2}$

**b)** $\\displaystyle\\frac{2}{y-3} + \\frac{5}{y+2}$

**c)** $\\displaystyle\\frac{1}{t-3} - \\frac{5}{t+4}$

**d)** $\\displaystyle\\frac{3}{w+2} - \\frac{2}{w-6}$

**e)** $\\displaystyle\\frac{1}{z+2} + \\frac{7}{z-7}$

For a repeated linear factor: $\\displaystyle\\frac{6x}{(x-2)(x+1)^2} = \\frac{A}{x-2} + \\frac{B}{x+1} + \\frac{C}{(x+1)^2}$

Multiply both sides by $(x-2)(x+1)^2$, substitute convenient values of $x$, then compare coefficients or substitute $x=0$ for any remaining unknown.

**Worked Example**

$\\displaystyle\\frac{4x+1}{(x-2)(x+1)^2} = \\frac{A}{x-2} + \\frac{B}{x+1} + \\frac{C}{(x+1)^2}$

$4x+1 = A(x+1)^2 + B(x-2)(x+1) + C(x-2)$

$x=2: \\;\\; 9 = 9A \\;\\Rightarrow\\; A = 1$

$x=-1: \\;\\; -3 = -3C \\;\\Rightarrow\\; C = 1$

$x=0: \\;\\; 1 = A - 2B - 2C \\;\\Rightarrow\\; 1 = 1 - 2B - 2 \\;\\Rightarrow\\; B = -1$

$\\displaystyle\\frac{4x+1}{(x-2)(x+1)^2} = \\frac{1}{x-2} - \\frac{1}{x+1} + \\frac{1}{(x+1)^2}$

## Question 2

Express each of the following as partial fractions.

**a)** $\\displaystyle\\frac{2x^2-x-3}{(x-2)(x-1)^2}$

**b)** $\\displaystyle\\frac{y^2-2y+8}{(y+2)(y-2)^2}$

**c)** $\\displaystyle\\frac{-3t^2+12t+7}{(t-3)(t+1)^2}$

**d)** $\\displaystyle\\frac{-3w^2+10w-11}{(w-2)(w-1)^2}$

Show Answers 

**a)** $\\displaystyle\\frac{3}{x-2} - \\frac{1}{x-1} + \\frac{2}{(x-1)^2}$

**b)** $\\displaystyle\\frac{1}{y+2} + \\frac{2}{(y-2)^2}$

**c)** $\\displaystyle\\frac{1}{t-3} - \\frac{4}{t+1} + \\frac{2}{(t+1)^2}$

**d)** $\\displaystyle\\frac{4}{(w-1)^2} - \\frac{3}{w-2}$

**Worked Example**

$\\displaystyle\\frac{2y+3}{y^2(y+1)} = \\frac{A}{y} + \\frac{B}{y^2} + \\frac{C}{y+1}$

$2y+3 = Ay(y+1) + B(y+1) + Cy^2$

$y=0: \\;\\; 3 = B \\;\\Rightarrow\\; B = 3$

$y=-1: \\;\\; 1 = C \\;\\Rightarrow\\; C = 1$

Compare $y^2$: $0 = A + C \\;\\Rightarrow\\; A = -1$

$\\displaystyle\\frac{2y+3}{y^2(y+1)} = -\\frac{1}{y} + \\frac{3}{y^2} + \\frac{1}{y+1}$

## Question 3

Express each of the following as partial fractions.

**a)** $\\displaystyle\\frac{2x^2-3}{(3-2x)(1-x)^2}$

**b)** $\\displaystyle\\frac{3y^2+17y+4}{y^2(y+4)}$

**c)** $\\displaystyle\\frac{t^2}{(t-2)(t-1)^2}$

**d)** $\\displaystyle\\frac{9w^2}{(2w+1)(w-1)^2}$

Show Answers 

**a)** $\\displaystyle\\frac{6}{3-2x} - \\frac{2}{1-x} - \\frac{1}{(1-x)^2}$

**b)** $\\displaystyle\\frac{1}{y^2} + \\frac{4}{y} - \\frac{1}{y+4}$

**c)** $\\displaystyle\\frac{4}{t-2} - \\frac{1}{(t-1)^2} - \\frac{3}{t-1}$

**d)** $\\displaystyle\\frac{4}{w-1} + \\frac{3}{(w-1)^2} + \\frac{1}{2w+1}$

If the degree of the numerator is $\\geq$ the degree of the denominator, first divide the numerator by the denominator: $\\displaystyle\\frac{f(x)}{g(x)} = Q(x) + \\frac{R(x)}{g(x)}$, where $Q(x)$ is the quotient and $R(x)$ the remainder.

**Worked Example**

$\\displaystyle\\frac{x^3+3x^2-5}{x+2}$

$x^3+3x^2-5 = (x+2)(x^2+x-2) - 1$

$\\displaystyle\\frac{x^3+3x^2-5}{x+2} = x^2+x-2 - \\frac{1}{x+2}$

## Question 4

For each of the following, find the constants $a$, $b$, $c$ and $d$ (or $a$, $b$, $c$ where only three appear).

**a)** $\\displaystyle\\frac{x^3+2x^2+3x-4}{x+1} \\equiv ax^2+bx+c+\\frac{d}{x+1}$

**b)** $\\displaystyle\\frac{2x^3+3x^2-4x+5}{x+3} \\equiv ax^2+bx+c+\\frac{d}{x+3}$

**c)** $\\displaystyle\\frac{2x^2+4x+5}{x^2-1} \\equiv a + \\frac{bx+c}{x^2-1}$

Show Answers 

**a)** $a=1,\\; b=1,\\; c=2,\\; d=-6$

**b)** $a=2,\\; b=-3,\\; c=5,\\; d=-10$

**c)** $a=2,\\; b=4,\\; c=7$

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