Repeated and quadratic factors · Partial fractions

In lesson 1 every bracket was different and linear, so each got one constant over it. Two other kinds of bracket come up in Further Maths: a bracket that is repeated, such as $(x + 1)^3$ , and a quadratic that won’t factorise, such as $x^2 + 2$ . Each needs its own kind of fraction:

Which fractions to writeOne fraction for each power of a repeated bracket; Bx + C over a quadratic

A repeated bracket

A single fraction $\dfrac{A}{(x + 1)^2}$ isn’t enough, because its top is only a number. Adding two fractions gives a top that can be any linear expression:

Add them: ${\frac{A}{x + 1} + \frac{B}{(x + 1)^2} = \frac{A(x + 1) + B}{(x + 1)^2}}$ .
The top, ${Ax + (A + B)}$ , can match any top $px + q$ .

So a bracket squared gets two fractions, one over each power, and a bracket cubed gets three.

To find the constants, multiply through as before. Only one value of $x$ makes the bracket zero, so it finds only the constant over the highest power. Compare coefficients to find the rest.

For example, to split $\dfrac{x + 4}{(x + 1)^2}$ :

Write ${\frac{A}{x + 1} + \frac{B}{(x + 1)^2}}$ .
Multiply through by $(x + 1)^2$ : ${x + 4 = A(x + 1) + B}$ .
Put $x = -1$ : ${3 = B}$ .
Compare the $x$ terms: ${1 = A}$ .
So ${\frac{x + 4}{(x + 1)^2} = \frac{1}{x + 1} + \frac{3}{(x + 1)^2}}$ .

The pieces are a new way of writing the same function. Slide $x$ and add them up:

The pieces add upSlide x; hide a piece

0.3331/(x + 1)0.3333/(x + 1)²0.667the pieces added0.667the fraction

x = 2

At x = 2, the pieces add up to 0.667, the value of the fraction. That happens at every x, because the partial fractions are the same function written another way.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q9 (a)

Resolve $\dfrac{x^2 + 1}{(x + 2)^3}$ into partial fractions.

Set up
- Three fractions, one over each power: ${\frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{(x + 2)^3}}$ .
- Multiply through by $(x + 2)^3$ : ${x^2 + 1 = A(x + 2)^2 + B(x + 2) + C}$ .
Think first. (x + 2)³ is repeated three times. How many fractions?
Make the bracket zero
- ${(-2)^2 + 1 = 0 + 0 + C}$ .
- So ${C = 5}$ .
Think first. Put x = −2.
Compare coefficients
- Expand: ${A(x^2 + 4x + 4) + Bx + 2B + C}$ .
- The $x^2$ terms: ${1 = A}$ .
- The $x$ terms: ${0 = 4A + B}$ , so ${B = -4}$ .
- Check the constants: ${4A + 2B + C = 4 - 8 + 5 = 1}$ ✓.
Think first. Expand A(x + 2)². Then match the x² terms and the x terms.
Answer
$\begin{aligned} \frac{x^2 + 1}{(x + 2)^3} = {} & \frac{1}{x + 2} - \frac{4}{(x + 2)^2} \\ & + \frac{5}{(x + 2)^3} \end{aligned}$

When there is a repeated bracket and a different one, use both ideas: substitute for the constants you can, then compare coefficients for the one that is left.

A shortcut: substitute for the bracket

When the whole bottom is one repeated bracket, such as $(x - 2)^4$ , put $u$ for the bracket. The top becomes a polynomial in $u$ , and dividing by $u^4$ splits it straight away.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q3

Using the substitution $u = x - 2$ , write $\dfrac{x^3 + 5}{(x - 2)^4}$ as an expression in terms of $u$ .

Using the answer in (a), express $\dfrac{x^3 + 5}{(x - 2)^4}$ in partial fractions.

Write x in terms of u
- ${u = x - 2}$ , so ${x = u + 2}$ .
Think first. u = x − 2, so x = ?
The top in terms of u
- ${x^3 = (u + 2)^3 = u^3 + 6u^2 + 12u + 8}$ .
- Add 5: ${x^3 + 5 = u^3 + 6u^2 + 12u + 13}$ .
- So the fraction is ${\frac{u^3 + 6u^2 + 12u + 13}{u^4}}$ .
Think first. Expand (u + 2)³.
Divide each term
- Split the fraction: ${\frac{u^3}{u^4} + \frac{6u^2}{u^4} + \frac{12u}{u^4} + \frac{13}{u^4}}$ .
- Cancel the powers of $u$ : ${\frac1u + \frac{6}{u^2} + \frac{12}{u^3} + \frac{13}{u^4}}$ .
Think first. Divide each term of the top by u⁴.
Put x back
$\begin{aligned} & \frac{1}{x - 2} + \frac{6}{(x - 2)^2} \\ & + \frac{12}{(x - 2)^3} + \frac{13}{(x - 2)^4} \end{aligned}$

A quadratic factor that won’t factorise

A quadratic such as $x^2 + 2$ has no real roots, so it can’t be split into brackets. Over it, the top can be linear, so write $Bx + C$ , not just a number. No value of $x$ makes it zero, so find $B$ and $C$ by comparing coefficients.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q10 (b)

A function $f$ is defined by $f(x) = \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 2)}$ . Express $f(x)$ in partial fractions.

Set up
- Write ${\frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2}}$ .
- Multiply through by $(x + 1)(x^2 + 2)$ . The left side becomes ${2x^2 - x + 3}$ .
- The right side becomes ${A(x^2 + 2) + (Bx + C)(x + 1)}$ .
Think first. What goes over x² + 2?
Make the linear bracket zero
- ${2 + 1 + 3 = A(1 + 2) + 0}$ .
- So ${3A = 6}$ , and ${A = 2}$ .
Think first. Put x = −1.
Compare coefficients
- Expand: ${A x^2 + 2A + Bx^2 + Bx + Cx + C}$ .
- The $x^2$ terms: ${2 = A + B}$ , so ${B = 0}$ .
- The constants: ${3 = 2A + C}$ , so ${C = -1}$ .
- Check the $x$ terms: ${B + C = -1}$ ✓.
Think first. Expand (Bx + C)(x + 1), then match the x² terms and the constants.
Answer
$f(x) = \frac{2}{x + 1} - \frac{1}{x^2 + 2}$

More: quadratic factors

WAEC 2009 · Paper 2 · Q9Express $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)}$ in partial fractions.

Why partial fractions matter

Each piece is easy to work with on its own. In integration (FM13), $\dfrac{x + 6}{(x + 1)^3}$ can’t be integrated as it stands, but its partial fractions can, term by term. Questions often ask for the partial fractions first and then use them.

More: partial fractions, then integrate

WAEC 2017 · Paper 2 · Q2Evaluate $\displaystyle\int_1^3 \frac{x - 1}{(x + 1)^2}\,dx$ (3 d.p.).

Your turn

WAEC 2016 · Paper 2 · Q9 (a)

(a)
Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions.

Worked solution (try it first)

(a)

Write $\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3}$ and multiply through: $x + 6 = A(x + 1)^2 + B(x + 1) + C$ .
Put $x = -1$ : $5 = C$ .
The $x^2$ terms: $0 = A$ .
The $x$ terms: $1 = 2A + B$ , so $B = 1$ .
So the answer is $\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}$ .

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