Algebraic Fractions with Quadratic Denominators

How to add and subtract algebraic fractions whose denominators are quadratic expressions: factorise every denominator, build the LCM from each factor to its highest power, and simplify the result.

Before you start

In lesson 4.4 you added fractions over the lowest common multiple (LCM) of their denominators, such as $\dfrac{1}{x - 2} + \dfrac{2}{x - 3} = \dfrac{3x - 7}{(x - 2)(x - 3)}\text{.}$

Now a denominator may be a quadratic expression, such as $x^2 - 9$ or $x^2 - 6x + 9$ . You cannot see its factors until you factorise it (chapter 3): $x^2 - 9 = {(x + 3)}{(x - 3)}$ and $x^2 - 6x + 9 = {(x - 3)}^2\text{.}$

Factorise the Denominators First

The LCM of algebraic denominators

1. Factorise every denominator completely.

2. List every different factor (number, letter or bracket).

3. Take each factor to the highest power it has in any one denominator (the greatest number of times it appears in any one denominator, as in lesson 4.4), and multiply them.

For example, ${(x + 3)}{(x - 3)}$ and ${(x - 3)}$ have the LCM ${(x + 3)}{(x - 3)}$ ; ${(x - 3)}$ and ${(x - 3)}^2$ have the LCM ${(x - 3)}^2$ .

Example 1: A Quadratic and a Linear Denominator

Express $\dfrac{1}{x^2 - 9} + \dfrac{2}{x - 3}$ as a single fraction in its simplest form.

Solution

Factorise: $x^2 - 9 = {(x + 3)}{(x - 3)}\text{.}$ The second denominator is ${(x - 3)}$ .

The factors are ${(x + 3)}$ and ${(x - 3)}$ , each to the power $1$ . LCM: ${(x + 3)}{(x - 3)}$ .

The first fraction already has this denominator. The second is missing ${(x + 3)}$ :

$\dfrac{1}{(x + 3)(x - 3)} + \dfrac{2(x + 3)}{(x + 3)(x - 3)} = \dfrac{1 + 2x + 6}{(x + 3)(x - 3)} = \dfrac{2x + 7}{(x + 3)(x - 3)}\text{.}$

$2x + 7$ shares no factor with the denominator, so this is the simplest form.

Check with $x = 4$ : $\dfrac{1}{7} + \dfrac{2}{1} = \dfrac{15}{7}\text{,}$ and $\dfrac{15}{7 \times 1} = \dfrac{15}{7}$ .

$\dfrac{2x + 7}{(x + 3)(x - 3)}$

Example 2: A Letter Factor as Well

Simplify $\dfrac{2}{x^2 + 3x} + \dfrac{1}{3x}$ .

Solution

Factorise: $x^2 + 3x = x{(x + 3)}$ . The second denominator is $3 \times x$ .

Factors: $3$ , $x$ and ${(x + 3)}$ . LCM: $3x{(x + 3)}$ .

The first fraction is missing $3$ ; the second is missing ${(x + 3)}$ :

$\dfrac{2 \times 3}{3x(x + 3)} + \dfrac{x + 3}{3x(x + 3)} = \dfrac{6 + x + 3}{3x(x + 3)} = \dfrac{x + 9}{3x(x + 3)}\text{.}$

Check with $x = 1$ : $\dfrac{2}{4} + \dfrac{1}{3} = \dfrac{5}{6}$ , and $\dfrac{10}{3 \times 4} = \dfrac{5}{6}$ .

$\dfrac{x + 9}{3x(x + 3)}$

A Repeated Factor

${(x - 3)}^2 = {(x - 3)}{(x - 3)}$ already contains ${(x - 3)}$ . So the LCM of ${(x - 3)}$ and ${(x - 3)}^2$ is ${(x - 3)}^2$ , the higher power. It is not ${(x - 3)}^3$ : that is a common multiple, but not the lowest.

The fraction over ${(x - 3)}$ is missing one factor ${(x - 3)}$ , so its numerator is multiplied by ${(x - 3)}$ .

Example 3: A Bracket and Its Square

Express $\dfrac{1}{x - 3} + \dfrac{2}{(x - 3)^2}$ as a single fraction.

Solution

LCM: ${(x - 3)}^2$ .

$\dfrac{1 \times (x - 3)}{(x - 3)^2} + \dfrac{2}{(x - 3)^2} = \dfrac{x - 3 + 2}{(x - 3)^2} = \dfrac{x - 1}{(x - 3)^2}\text{.}$

Check with $x = 4$ : $\dfrac{1}{1} + \dfrac{2}{1} = 3$ , and $\dfrac{3}{1^2} = 3$ . With $x = 5$ : $\dfrac{1}{2} + \dfrac{2}{4} = 1$ , and $\dfrac{4}{4} = 1$ .

$\dfrac{x - 1}{(x - 3)^2}$

Example 4: A Perfect-Square Denominator

Simplify $\dfrac{1}{x^2 - 6x + 9} - \dfrac{1}{x - 3}\text{.}$

Solution

Factorise: $x^2 - 6x + 9 = {(x - 3)}^2\text{,}$ a perfect square. LCM: ${(x - 3)}^2$ .

$\dfrac{1}{(x - 3)^2} - \dfrac{x - 3}{(x - 3)^2} = \dfrac{1 - (x - 3)}{(x - 3)^2} = \dfrac{1 - x + 3}{(x - 3)^2} = \dfrac{4 - x}{(x - 3)^2}\text{.}$

Check with $x = 5$ : $\dfrac{1}{4} - \dfrac{1}{2} = -\dfrac{1}{4}\text{,}$ and $\dfrac{4 - 5}{4} = -\dfrac{1}{4}$ .

$\dfrac{4 - x}{(x - 3)^2}$

Two Quadratic Denominators

When both denominators are quadratics, factorise both. If they share a bracket, the LCM takes it only once.

Example 5: Denominators Sharing a Factor

Simplify $\dfrac{1}{x^2 - x - 2} + \dfrac{1}{x^2 - 4}\text{.}$

Solution

Factorise: $x^2 - x - 2 = {(x - 2)}{(x + 1)}$ (product $-2$ , sum $-1$ ) and $x^2 - 4 = {(x + 2)}{(x - 2)}\text{.}$

Factors: ${(x - 2)}$ , ${(x + 1)}$ , ${(x + 2)}$ . LCM: ${(x - 2)}{(x + 1)}{(x + 2)}\text{.}$

The first fraction is missing ${(x + 2)}$ ; the second is missing ${(x + 1)}$ :

$\dfrac{(x + 2) + (x + 1)}{(x - 2)(x + 1)(x + 2)} = \dfrac{2x + 3}{(x - 2)(x + 1)(x + 2)}\text{.}$

Check with $x = 3$ : $\dfrac{1}{4} + \dfrac{1}{5} = \dfrac{9}{20}\text{,}$ and $\dfrac{9}{1 \times 4 \times 5} = \dfrac{9}{20}\text{.}$

$\dfrac{2x + 3}{(x - 2)(x + 1)(x + 2)}$

Why not just multiply the two denominators?

Multiplying ${(x - 2)}{(x + 1)}$ by ${(x + 2)}{(x - 2)}$ also gives a common denominator, and the answer you get is still right. But it has ${(x - 2)}$ twice, so the numerator then has a factor ${(x - 2)}$ too, and you must cancel it at the end. Using the LCM avoids the extra work.

Subtracting, and Results That Cancel

Example 6: Subtracting with a Quadratic Denominator

Simplify $\dfrac{3}{x^2 - 1} - \dfrac{1}{x - 1}\text{.}$

Solution

$x^2 - 1 = {(x + 1)}{(x - 1)}\text{.}$ LCM: ${(x + 1)}{(x - 1)}$ . The second fraction is missing ${(x + 1)}$ .

$\dfrac{3 - (x + 1)}{(x + 1)(x - 1)} = \dfrac{3 - x - 1}{(x + 1)(x - 1)} = \dfrac{2 - x}{(x + 1)(x - 1)}\text{.}$

The bracket round ${(x + 1)}$ makes the minus sign reach both terms.

Check with $x = 3$ : $\dfrac{3}{8} - \dfrac{1}{2} = -\dfrac{1}{8}\text{,}$ and $\dfrac{-1}{4 \times 2} = -\dfrac{1}{8}\text{.}$

$\dfrac{2 - x}{(x + 1)(x - 1)}$

Example 7: A Result That Simplifies

Simplify $\dfrac{1}{x - 2} - \dfrac{4}{x^2 - 4}\text{.}$

Solution

$x^2 - 4 = {(x + 2)}{(x - 2)}\text{.}$ LCM: ${(x + 2)}{(x - 2)}$ .

$\dfrac{(x + 2) - 4}{(x + 2)(x - 2)} = \dfrac{x - 2}{(x + 2)(x - 2)}\text{.}$

Always look at the numerator again: here it is the factor ${(x - 2)}$ , which cancels: $\dfrac{1}{x + 2}$ , for $x \neq \pm 2$ .

Check with $x = 3$ : $1 - \dfrac{4}{5} = \dfrac{1}{5}$ , and $\dfrac{1}{5}$ .

$\dfrac{1}{x + 2}$

Example 8: A Bracket Written Backwards

Simplify $\dfrac{1}{x^2 - 9} + \dfrac{1}{3 - x}\text{.}$

Solution

$3 - x = -{(x - 3)}$ , so $\dfrac{1}{3 - x} = -\dfrac{1}{x - 3}\text{.}$ The sum becomes $\dfrac{1}{(x + 3)(x - 3)} - \dfrac{1}{x - 3}\text{.}$

LCM: ${(x + 3)}{(x - 3)}$ . $\dfrac{1 - (x + 3)}{(x + 3)(x - 3)} = \dfrac{-x - 2}{(x + 3)(x - 3)} = -\dfrac{x + 2}{(x + 3)(x - 3)}\text{.}$

Check with $x = 4$ : $\dfrac{1}{7} + \dfrac{1}{-1} = -\dfrac{6}{7}\text{,}$ and $-\dfrac{6}{7 \times 1} = -\dfrac{6}{7}\text{.}$

$-\dfrac{x + 2}{(x + 3)(x - 3)}$

Where is it undefined?

Read the values from the factorised LCM. In Example 5 the expression is undefined when $x = 2$ , $x = -1$ or $x = -2$ . Even when a factor cancels at the end, as in Example 7, its value stays excluded: $\dfrac{1}{x - 2} - \dfrac{4}{x^2 - 4}$ has no value at $x = 2$ .

Combining Operations

Example 9: Subtract, Then Divide

Work out $\dfrac{1}{x - 1} - \dfrac{1}{x + 1}$ , then divide the result by $\dfrac{2}{x^2 - 1}$ , and simplify.

Solution

Work out the difference first. LCM: ${(x - 1)}{(x + 1)}$ .

$\dfrac{(x + 1) - (x - 1)}{(x - 1)(x + 1)} = \dfrac{2}{(x - 1)(x + 1)}\text{.}$

Then divide, multiplying by the reciprocal (lesson 4.3), with $x^2 - 1 = {(x + 1)}{(x - 1)}\text{:}$

$\dfrac{2}{(x - 1)(x + 1)} \times \dfrac{(x + 1)(x - 1)}{2} = 1\text{,}$ for $x \neq \pm 1$ .

Check with $x = 2$ : the difference is $1 - \dfrac{1}{3} = \dfrac{2}{3}$ , and $\dfrac{2}{3} \div \dfrac{2}{3} = 1$ .

$1$

Common Mistakes

Not factorising the quadratic denominator first, and so missing the factor it shares with the other denominator.
Taking ${(x - 3)}^3$ as the LCM of ${(x - 3)}$ and ${(x - 3)}^2$ . The highest power, ${(x - 3)}^2$ , is enough.
Forgetting to multiply the numerator: over ${(x - 3)}^2$ , the fraction $\dfrac{1}{x - 3}$ becomes $\dfrac{x - 3}{(x - 3)^2}$ , not $\dfrac{1}{(x - 3)^2}$ .
Applying the minus to the first term only: $3 - {(x + 1)} = 2 - x$ , not $4 + x$ .
Stopping before the end: if the numerator factorises, see whether a factor cancels with the denominator.

Practice

Factorise every denominator first. Give each answer as a single fraction in its simplest form, with the denominator factorised.

Question 1

Find the LCM of each pair of denominators. (a) $x^2 - 4$ and $x + 2$ (b) $x^2 + 2x + 1$ and $x^2 - 1$ (c) $2x - 6$ and $x^2 - 9$ (d) $x^2 + 5x$ and $x^2 - 25$

ShowHide solution

(a) ${(x + 2)}{(x - 2)}$ and ${(x + 2)}$ : LCM ${(x + 2)}{(x - 2)}$ .

(b) ${(x + 1)}^2$ and ${(x + 1)}{(x - 1)}$ : the highest power of ${(x + 1)}$ is $2$ . LCM ${(x + 1)}^2{(x - 1)}$ .

(d) $x{(x + 5)}$ and ${(x + 5)}{(x - 5)}$ : LCM $x{(x + 5)}{(x - 5)}$ .

(a) ${(x + 2)}{(x - 2)}$ (b) ${(x + 1)}^2{(x - 1)}$ (c) $2{(x + 3)}{(x - 3)}$ (d) $x{(x + 5)}{(x - 5)}$

Question 2

Simplify (a) $\dfrac{3}{x^2 - 4} + \dfrac{1}{x + 2}$ (b) $\dfrac{5}{x^2 - 16} + \dfrac{2}{x - 4}$

ShowHide solution

(a) LCM ${(x + 2)}{(x - 2)}$ : $\dfrac{3 + (x - 2)}{(x + 2)(x - 2)} = \dfrac{x + 1}{(x + 2)(x - 2)}\text{.}$

(b) LCM ${(x + 4)}{(x - 4)}$ : $\dfrac{5 + 2(x + 4)}{(x + 4)(x - 4)} = \dfrac{2x + 13}{(x + 4)(x - 4)}\text{.}$

(a) $\dfrac{x + 1}{(x + 2)(x - 2)}$ (b) $\dfrac{2x + 13}{(x + 4)(x - 4)}$

Question 3

Simplify (a) $\dfrac{3}{x + 2} + \dfrac{1}{(x + 2)^2}$ (b) $\dfrac{2}{(x - 1)^2} - \dfrac{1}{x - 1}$

ShowHide solution

(a) LCM ${(x + 2)}^2$ : $\dfrac{3(x + 2) + 1}{(x + 2)^2} = \dfrac{3x + 7}{(x + 2)^2}\text{.}$

(b) LCM ${(x - 1)}^2$ : $\dfrac{2 - (x - 1)}{(x - 1)^2} = \dfrac{3 - x}{(x - 1)^2}\text{.}$

Check (b) with $x = 2$ : $2 - 1 = 1$ , and $\dfrac{1}{1} = 1$ .

(a) $\dfrac{3x + 7}{(x + 2)^2}$ (b) $\dfrac{3 - x}{(x - 1)^2}$

Question 4

Simplify (a) $\dfrac{x}{x^2 + 4x + 4} + \dfrac{2}{x + 2}$ (b) $\dfrac{1}{x^2 + 3x + 2} + \dfrac{1}{x^2 + 4x + 3}$

ShowHide solution

(a) $x^2 + 4x + 4 = {(x + 2)}^2\text{.}$ $\dfrac{x + 2(x + 2)}{(x + 2)^2} = \dfrac{3x + 4}{(x + 2)^2}\text{.}$

(b) ${(x + 1)}{(x + 2)}$ and ${(x + 1)}{(x + 3)}$ ; LCM ${(x + 1)}{(x + 2)}{(x + 3)}\text{.}$

$\dfrac{(x + 3) + (x + 2)}{(x + 1)(x + 2)(x + 3)} = \dfrac{2x + 5}{(x + 1)(x + 2)(x + 3)}\text{.}$

(a) $\dfrac{3x + 4}{(x + 2)^2}$ (b) $\dfrac{2x + 5}{(x + 1)(x + 2)(x + 3)}$

Question 5

Simplify (a) $\dfrac{x}{x^2 - 1} - \dfrac{1}{x + 1}$ (b) $\dfrac{6}{x^2 - 9} - \dfrac{1}{x - 3}$ (c) $\dfrac{2}{x + 3} + \dfrac{12}{x^2 - 9}$

ShowHide solution

(a) $\dfrac{x - (x - 1)}{(x + 1)(x - 1)} = \dfrac{1}{(x + 1)(x - 1)}\text{.}$

(b) $x^2 - 9 = {(x + 3)}{(x - 3)}\text{,}$ which is the LCM. $\dfrac{6 - (x + 3)}{(x + 3)(x - 3)} = \dfrac{3 - x}{(x + 3)(x - 3)}\text{.}$ Since $3 - x = -{(x - 3)}$ , this is $-\dfrac{1}{x + 3}$ .

Check (b) with $x = 4$ : $\dfrac{6}{7} - 1 = -\dfrac{1}{7}$ , and $-\dfrac{1}{4 + 3} = -\dfrac{1}{7}\text{.}$

(c) $\dfrac{2(x - 3) + 12}{(x + 3)(x - 3)} = \dfrac{2x + 6}{(x + 3)(x - 3)} = \dfrac{2(x + 3)}{(x + 3)(x - 3)} = \dfrac{2}{x - 3}\text{.}$

Check (c) with $x = 4$ : $\dfrac{2}{7} + \dfrac{12}{7} = 2$ , and $\dfrac{2}{1} = 2$ .

(a) $\dfrac{1}{(x + 1)(x - 1)}$ (b) $-\dfrac{1}{x + 3}$ (c) $\dfrac{2}{x - 3}$

Question 6

Simplify $\dfrac{2}{x^2 - 4} + \dfrac{1}{2 - x}\text{.}$

ShowHide solution

$\dfrac{1}{2 - x} = -\dfrac{1}{x - 2}\text{.}$ LCM ${(x - 2)}{(x + 2)}$ .

$\dfrac{2 - (x + 2)}{(x - 2)(x + 2)} = \dfrac{-x}{(x - 2)(x + 2)}\text{.}$

Check with $x = 3$ : $\dfrac{2}{5} - 1 = -\dfrac{3}{5}$ , and $\dfrac{-3}{1 \times 5} = -\dfrac{3}{5}\text{.}$

$-\dfrac{x}{(x - 2)(x + 2)}$

Question 7

Simplify $\dfrac{3}{x^2 - 2x} - \dfrac{1}{2x}$ .

ShowHide solution

$x^2 - 2x = x{(x - 2)}$ . LCM $2x{(x - 2)}$ .

$\dfrac{3 \times 2 - (x - 2)}{2x(x - 2)} = \dfrac{6 - x + 2}{2x(x - 2)} = \dfrac{8 - x}{2x(x - 2)}\text{.}$

$\dfrac{8 - x}{2x(x - 2)}$

Question 8

Simplify $\dfrac{1}{x^2 + x} + \dfrac{1}{x^2 - x}\text{,}$ and state the values of $x$ for which it is undefined.

ShowHide solution

$x{(x + 1)}$ and $x{(x - 1)}$ ; LCM $x{(x + 1)}{(x - 1)}$ .

$\dfrac{(x - 1) + (x + 1)}{x(x + 1)(x - 1)} = \dfrac{2x}{x(x + 1)(x - 1)} = \dfrac{2}{(x + 1)(x - 1)}\text{.}$

The original is undefined when $x = 0$ , $x = -1$ or $x = 1$ . The value $x = 0$ stays excluded although the $x$ cancelled.

$\dfrac{2}{(x + 1)(x - 1)}$ ; undefined when $x = 0$ , $1$ or $-1$

Question 9

(a) Kumar works out $\dfrac{1}{x^2 - 9} + \dfrac{1}{x - 3}$ over the common denominator ${(x^2 - 9)}{(x - 3)}$ and gets $\dfrac{x^2 + x - 12}{(x^2 - 9)(x - 3)}\text{.}$ Is he right? Is his answer in its simplest form? (b) Ahmad writes $\dfrac{x + 5}{x^2 - 1} - \dfrac{2}{x + 1} = \dfrac{x + 5 - 2x - 2}{(x + 1)(x - 1)}\text{.}$ Find his error.

ShowHide solution

(a) His working is right, but he used a common multiple that is not the lowest. $x^2 + x - 12 = {(x + 4)}{(x - 3)}\text{,}$ and ${(x - 3)}$ cancels: $\dfrac{x + 4}{(x + 3)(x - 3)}\text{.}$ With the LCM ${(x + 3)}{(x - 3)}$ : $\dfrac{1 + (x + 3)}{(x + 3)(x - 3)} = \dfrac{x + 4}{(x + 3)(x - 3)}$ at once.

(b) The second fraction needs $2{(x - 1)} = 2x - 2$ on top, and the minus makes it $-2x + 2$ . Correct: $\dfrac{x + 5 - 2x + 2}{(x + 1)(x - 1)} = \dfrac{7 - x}{(x + 1)(x - 1)}\text{.}$ Check with $x = 2$ : $\dfrac{7}{3} - \dfrac{2}{3} = \dfrac{5}{3}$ , and $\dfrac{5}{3}$ .

(a) Right, but not simplest: $\dfrac{x + 4}{(x + 3)(x - 3)}$ (b) $\dfrac{7 - x}{(x + 1)(x - 1)}$

Question 10

Work out $\dfrac{1}{x - 2} + \dfrac{1}{x + 2}$ , then multiply the result by $\dfrac{x^2 - 4}{x}$ , and simplify.

ShowHide solution

The sum: $\dfrac{(x + 2) + (x - 2)}{(x - 2)(x + 2)} = \dfrac{2x}{(x - 2)(x + 2)}\text{.}$

Multiply: $\dfrac{2x}{(x - 2)(x + 2)} \times \dfrac{(x + 2)(x - 2)}{x} = 2\text{,}$ for $x \neq 0$ , $\pm 2$ .

$2$

Question 11

A sum of $24 is shared equally among ${(x - 2)}$ pupils, and a sum of $48 is shared equally among ${(x^2 - 4)}$ pupils. (a) How much more does each pupil in the first group get? Give a single fraction. (b) Find this amount when $x = 6$ .

ShowHide solution

(a) $\dfrac{24}{x - 2} - \dfrac{48}{x^2 - 4}\text{.}$ LCM ${(x - 2)}{(x + 2)}$ : $\dfrac{24(x + 2) - 48}{(x - 2)(x + 2)} = \dfrac{24x}{(x - 2)(x + 2)}$ dollars.

(b) $\dfrac{144}{4 \times 8} = 4.5$ , that is, $4.50. Check: $24 ÷ 4 = $6 and $48 ÷ 32 = $1.50; the difference is $4.50.

(a) $\dfrac{24x}{(x - 2)(x + 2)}$ dollars (b) $4.50

Question 12

A boat goes $6$ km up a river at ${(x - 3)}$ km/h and comes back $6$ km at ${(x + 3)}$ km/h. (a) Find the total time as a single fraction. (b) How much longer does the trip up the river take than the trip down? (c) Find both answers when $x = 9$ .

ShowHide solution

(a) $\dfrac{6}{x - 3} + \dfrac{6}{x + 3} = \dfrac{6(x + 3) + 6(x - 3)}{(x - 3)(x + 3)} = \dfrac{12x}{(x - 3)(x + 3)}$ hours.

(b) $\dfrac{6}{x - 3} - \dfrac{6}{x + 3} = \dfrac{6(x + 3) - 6(x - 3)}{(x - 3)(x + 3)} = \dfrac{36}{(x - 3)(x + 3)}$ hours.

(c) $\dfrac{108}{6 \times 12} = 1.5$ hours and $\dfrac{36}{72} = 0.5$ hours. Check: $6 \div 6 = 1$ hour up and $6 \div 12 = 0.5$ hours down.

(a) $\dfrac{12x}{(x - 3)(x + 3)}$ h (b) $\dfrac{36}{(x - 3)(x + 3)}$ h (c) $1\tfrac{1}{2}$ h and $\tfrac{1}{2}$ h

Key Points

Factorise every denominator before you look for the LCM.
The LCM takes each factor to the highest power it has in any one denominator: ${(x - 3)}$ and ${(x - 3)}^2$ give ${(x - 3)}^2$ .
Multiply each numerator by the factors its denominator is missing.
Put brackets round what is subtracted, then expand.
Leave the denominator factorised, and see whether a factor of the numerator cancels.
The expression is undefined where any factor of the LCM is zero, even if that factor cancels.