Simplifying Algebraic Fractions by Factorising

How to simplify an algebraic fraction whose common factors are hidden: factorise the numerator and the denominator completely, cancel the common brackets, and state where the original fraction is undefined.

Before you start

In lesson 4.1 you cancelled factors you could see, such as $\dfrac{4(x + 2)^2}{6x(x + 2)} = \dfrac{2(x + 2)}{3x}\text{.}$ Most fractions do not come with their factors on show: $\dfrac{x^2 - 9}{2x + 6}$ has terms, not factors, on top and below, so nothing can be cancelled yet.

Chapter 3 gave you the tools to make the factors appear. Here they are again.

Factorise First, Then Cancel

The ways to factorise (chapter 3)

Common factor: $x^2 - 3x = x{(x - 3)}$ ; $3x - 9 = 3{(x - 3)}$ .

Difference of two squares: $a^2 - b^2 = {(a + b)}{(a - b)}\text{,}$ so $x^2 - 9 = {(x + 3)}{(x - 3)}\text{.}$

Perfect squares: $a^2 + 2ab + b^2 = {(a + b)}^2$ and $a^2 - 2ab + b^2 = {(a - b)}^2\text{,}$ so $x^2 - 6x + 9 = {(x - 3)}^2\text{.}$

Quadratic $x^2 + bx + c$ : two numbers with product $c$ and sum $b$ , so $x^2 - 5x + 6 = {(x - 2)}{(x - 3)}\text{.}$

Quadratic $ax^2 + bx + c$ : two numbers with product $ac$ and sum $b$ ; split the middle term and group.

Grouping (four terms): $ax - ay + 2x - 2y = a{(x - y)} + 2{(x - y)} = {(a + 2)}{(x - y)}\text{.}$

Simplifying by factorising

1. Factorise the numerator completely, and the denominator completely. Take out a common factor first.

2. Cancel each factor (a number, a letter or a whole bracket) that appears in both.

3. State the values for which the original fraction is undefined: they make a factor of the original denominator zero.

Example 1: A Common Factor in the Numerator or the Denominator

Simplify (a) $\dfrac{a^2 + 4ab^2}{3ab}$ (b) $\dfrac{3t}{t^2 - 2t}$ , and state the values of $t$ for which (b) is undefined.

Solution

(a) The numerator has the common factor $a$ : $a^2 + 4ab^2 = a{(a + 4b^2)}\text{.}$

$\dfrac{a(a + 4b^2)}{3ab} = \dfrac{\cancel{a}(a + 4b^2)}{3\cancel{a}b} = \dfrac{a + 4b^2}{3b}\text{.}$ The bracket ${(a + 4b^2)}$ stays whole: its $a$ is a term inside it, not a factor of it.

(b) The denominator has the common factor $t$ : $t^2 - 2t = t{(t - 2)}$ .

$\dfrac{3t}{t(t - 2)} = \dfrac{3}{t - 2}\text{.}$ The original denominator $t{(t - 2)}$ is $0$ when $t = 0$ or $t = 2$ .

Check (a) with $a = 2$ , $b = 1$ : $\dfrac{4 + 8}{6} = 2$ and $\dfrac{2 + 4}{3} = 2$ .

(a) $\dfrac{a + 4b^2}{3b}$ (b) $\dfrac{3}{t - 2}$ , undefined when $t = 0$ or $t = 2$

Example 2: The Same Bracket Top and Bottom

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

Solution

Numerator: $x^2 - 3x = x{(x - 3)}$ . Denominator: $3x - 9 = 3{(x - 3)}$ .

$\dfrac{x\cancel{(x - 3)}}{3\cancel{(x - 3)}} = \dfrac{x}{3}\text{,}$ for $x \neq 3$ .

Check with $x = 4$ : $\dfrac{16 - 12}{12 - 9} = \dfrac{4}{3}\text{,}$ and $\dfrac{4}{3}$ .

At $x = 3$ the answer $\dfrac{x}{3}$ is $1$ , but the original is $\dfrac{0}{0}$ , which has no value. So the two are equal for every $x$ except $3$ .

$\dfrac{x}{3}$ , for $x \neq 3$

Using the Identities

Example 3: Difference of Two Squares

Simplify (a) $\dfrac{x^2 - 9}{2x + 6}$ (b) $\dfrac{2m^2 - 4m}{m^2 - 4}$ , stating the values for which each original fraction is undefined.

Solution

(a) $x^2 - 9 = x^2 - 3^2 = {(x + 3)}{(x - 3)}\text{,}$ and $2x + 6 = 2{(x + 3)}$ .

$\dfrac{\cancel{(x + 3)}(x - 3)}{2\cancel{(x + 3)}} = \dfrac{x - 3}{2}\text{.}$ The original is undefined when $2x + 6 = 0$ , that is, $x = -3$ .

(b) $2m^2 - 4m = 2m{(m - 2)}\text{,}$ and $m^2 - 4 = {(m + 2)}{(m - 2)}\text{.}$

$\dfrac{2m\cancel{(m - 2)}}{(m + 2)\cancel{(m - 2)}} = \dfrac{2m}{m + 2}\text{.}$ The original is undefined when $m = -2$ or $m = 2$ .

Check (a) with $x = 4$ : $\dfrac{7}{14} = \dfrac{1}{2}$ and $\dfrac{4 - 3}{2} = \dfrac{1}{2}$ . Check (b) with $m = 3$ : $\dfrac{18 - 12}{9 - 4} = \dfrac{6}{5}$ and $\dfrac{6}{5}$ .

(a) $\dfrac{x - 3}{2}$ , for $x \neq -3$ (b) $\dfrac{2m}{m + 2}$ , for $m \neq \pm 2$

Example 4: A Perfect Square

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

Solution

The numerator is a perfect square: $9 = 3^2$ and $6x = 2 \times x \times 3$ , so $x^2 - 6x + 9 = {(x - 3)}^2 = {(x - 3)}{(x - 3)}\text{.}$

The denominator is a difference of two squares: ${(x + 3)}{(x - 3)}$ .

$\dfrac{(x - 3)\cancel{(x - 3)}}{(x + 3)\cancel{(x - 3)}} = \dfrac{x - 3}{x + 3}\text{,}$ for $x \neq \pm 3$ .

Check with $x = 4$ : $\dfrac{16 - 24 + 9}{16 - 9} = \dfrac{1}{7}\text{,}$ and $\dfrac{1}{7}$ .

$\dfrac{x - 3}{x + 3}$ , for $x \neq 3$ and $x \neq -3$

Quadratic Expressions Top and Bottom

Example 5: Two Quadratics

Simplify (a) $\dfrac{x^2 - 5x + 6}{x^2 - 4}$ (b) $\dfrac{x^2 + x - 6}{x^2 + 5x + 6}$

Solution

(a) For $x^2 - 5x + 6$ : product $6$ , sum $-5$ , so $-2$ and $-3$ : ${(x - 2)}{(x - 3)}$ . And $x^2 - 4 = {(x + 2)}{(x - 2)}\text{.}$

$\dfrac{\cancel{(x - 2)}(x - 3)}{(x + 2)\cancel{(x - 2)}} = \dfrac{x - 3}{x + 2}\text{,}$ for $x \neq \pm 2$ .

(b) For $x^2 + x - 6$ : product $-6$ , sum $1$ , so $3$ and $-2$ : ${(x + 3)}{(x - 2)}$ . For $x^2 + 5x + 6$ : product $6$ , sum $5$ , so $2$ and $3$ : ${(x + 2)}{(x + 3)}$ .

$\dfrac{\cancel{(x + 3)}(x - 2)}{(x + 2)\cancel{(x + 3)}} = \dfrac{x - 2}{x + 2}\text{,}$ for $x \neq -2$ and $x \neq -3$ .

Check (a) with $x = 4$ : $\dfrac{2}{12} = \dfrac{1}{6}$ and $\dfrac{1}{6}$ . Check (b) with $x = 1$ : $\dfrac{-4}{12} = -\dfrac{1}{3}$ and $\dfrac{-1}{3}$ .

(a) $\dfrac{x - 3}{x + 2}$ , for $x \neq \pm 2$ (b) $\dfrac{x - 2}{x + 2}$ , for $x \neq -2$ and $x \neq -3$

Example 6: A Quadratic with $a \neq 1$

Simplify $\dfrac{2x^2 + 5x - 3}{4x^2 - 1}\text{.}$

Solution

Numerator: $ac = 2 \times {(-3)} = -6\text{,}$ and the two numbers with product $-6$ and sum $5$ are $6$ and $-1$ .

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

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

$\dfrac{\cancel{(2x - 1)}(x + 3)}{(2x + 1)\cancel{(2x - 1)}} = \dfrac{x + 3}{2x + 1}\text{,}$ for $x \neq \pm\dfrac{1}{2}$ .

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

$\dfrac{x + 3}{2x + 1}$ , for $x \neq \pm\dfrac{1}{2}$

Example 7: Two Letters

Simplify $\dfrac{x^2 - 3xy - 4y^2}{3x^2 - 12xy}\text{.}$

Solution

Numerator: treat it like $x^2 - 3x - 4$ with $y$ beside each number. Product $-4$ , sum $-3$ : $-4$ and $1$ . So $x^2 - 3xy - 4y^2 = {(x - 4y)}{(x + y)}\text{.}$

Denominator: $3x^2 - 12xy = 3x{(x - 4y)}\text{.}$

$\dfrac{\cancel{(x - 4y)}(x + y)}{3x\cancel{(x - 4y)}} = \dfrac{x + y}{3x}\text{,}$ for $x \neq 0$ and $x \neq 4y$ .

Check with $x = 1$ , $y = 1$ : $\dfrac{1 - 3 - 4}{3 - 12} = \dfrac{-6}{-9} = \dfrac{2}{3}\text{,}$ and $\dfrac{2}{3}$ .

$\dfrac{x + y}{3x}$ , for $x \neq 0$ and $x \neq 4y$

Factorising by Grouping

Example 8: Four Terms on Top

Simplify $\dfrac{ax - ay + 2x - 2y}{x^2 - y^2}\text{.}$

Solution

Group the numerator in pairs: $a{(x - y)} + 2{(x - y)} = {(a + 2)}{(x - y)}\text{.}$

Denominator: $x^2 - y^2 = {(x + y)}{(x - y)}\text{.}$

$\dfrac{(a + 2)\cancel{(x - y)}}{(x + y)\cancel{(x - y)}} = \dfrac{a + 2}{x + y}\text{,}$ for $x \neq y$ and $x \neq -y$ .

Check with $a = 1$ , $x = 2$ , $y = 1$ : $\dfrac{2 - 1 + 4 - 2}{4 - 1} = \dfrac{3}{3} = 1\text{,}$ and $\dfrac{3}{3} = 1$ .

$\dfrac{a + 2}{x + y}$ , for $x \neq \pm y$

Brackets That Are Negatives of Each Other

Look at $x - 3$ and $3 - x$ . When $x = 5$ they are $2$ and $-2$ ; when $x = 1$ they are $-2$ and $2$ . One is always the negative of the other: $3 - x = -{(x - 3)}$ .

So when a fraction has ${(x - 3)}$ on one side and ${(3 - x)}$ on the other, write ${(3 - x)}$ as $-{(x - 3)}$ . The brackets then cancel and leave the factor $-1$ .

Reversing a bracket

$b - a = -{(a - b)}$ , so $\dfrac{a - b}{b - a} = -1$ (for $a \neq b$ ).

Example 9: Using $b - a = -{(a - b)}$

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

Solution

(a) $\dfrac{x - 3}{-(x - 3)} = -1\text{,}$ for $x \neq 3$ .

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

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

Check (b) with $x = 1$ : $\dfrac{4 - 1}{1 - 2} = -3$ , and $-\dfrac{3}{1} = -3$ .

(a) $-1$ , for $x \neq 3$ (b) $-\dfrac{x + 2}{x}$ , for $x \neq 0$ and $x \neq 2$

When Nothing Cancels, and Finishing the Job

If, after factorising completely, the numerator and the denominator have no common factor, the fraction is already in its simplest form. And if a bracket can still be factorised, you have not finished: a common factor may be hidden inside it.

Example 10: Simplify If Possible

Simplify where possible. (a) $\dfrac{x + 3}{x + 5}$ (b) $\dfrac{x^2 + 4}{x + 2}$ (c) $\dfrac{2x^2 - 8}{x - 2}$

Solution

(a) Neither $x + 3$ nor $x + 5$ factorises, and they are different. Nothing cancels: each $x$ is a term, not a factor. Testing $x = 1$ : $\dfrac{4}{6} = \dfrac{2}{3}$ , which is not $\dfrac{3}{5}$ , so cancelling the $x$ would be wrong.

(b) $x^2 + 4$ is a sum of two squares, not a difference, and does not factorise. So ${(x + 2)}$ is not a factor of the numerator, and the fraction cannot be simplified.

(c) $2x^2 - 8 = 2{(x^2 - 4)}$ . Do not stop there: $x^2 - 4 = {(x + 2)}{(x - 2)}\text{.}$ So $\dfrac{2(x + 2)\cancel{(x - 2)}}{\cancel{(x - 2)}} = 2{(x + 2)}\text{,}$ for $x \neq 2$ .

Check (c) with $x = 3$ : $\dfrac{18 - 8}{1} = 10$ , and $2 \times 5 = 10$ .

(a) and (b) cannot be simplified. (c) $2{(x + 2)}$ , for $x \neq 2$

Common Mistakes

Cancelling before factorising: $\dfrac{x^2 - 4}{x - 2} = x - 2$ "because $x^2 \div x = x$ and $4 \div 2 = 2$ " is wrong. $x^2 - 4 = {(x + 2)}{(x - 2)}\text{,}$ so the answer is $x + 2$ . Test $x = 3$ : $\dfrac{5}{1} = 5$ , not $1$ .
Cancelling terms: $\dfrac{x + 3}{x + 5} = \dfrac{3}{5}$ is wrong ( $x = 1$ gives $\dfrac{2}{3}$ ).
Stopping too soon: $\dfrac{2(x^2 - 4)}{x - 2}$ is not finished until $x^2 - 4$ is factorised.
Missing a reversed bracket: ${(3 - x)}$ and ${(x - 3)}$ cancel to $-1$ , not $1$ .
Factorising a sum of two squares: $x^2 + 4 \neq {(x + 2)}{(x + 2)}\text{,}$ which is $x^2 + 4x + 4$ .
Taking the excluded values from the answer. In $\dfrac{x^2 - 3x}{3x - 9} = \dfrac{x}{3}\text{,}$ the value $x = 3$ is still excluded.

Practice

Factorise completely before you cancel. Check each answer by putting a number in for each letter on both sides.

Question 1

Simplify (a) $\dfrac{h^2 + 7hk}{5hk}$ (b) $\dfrac{6x + 12}{x^2 + 2x}$ (c) $\dfrac{5y - 15}{10y - 30}$

ShowHide solution

(a) $\dfrac{h(h + 7k)}{5hk} = \dfrac{h + 7k}{5k}$

(b) $\dfrac{6(x + 2)}{x(x + 2)} = \dfrac{6}{x}$

(a) $\dfrac{h + 7k}{5k}$ (b) $\dfrac{6}{x}$ (c) $\dfrac{1}{2}$

Question 2

Simplify, stating the values for which each original fraction is undefined. (a) $\dfrac{x^2 - 16}{3x - 12}$ (b) $\dfrac{3v^2 - 9v}{v^2 - 9}$

ShowHide solution

(a) $\dfrac{(x + 4)(x - 4)}{3(x - 4)} = \dfrac{x + 4}{3}\text{.}$ Undefined when $3x - 12 = 0$ : $x = 4$ .

(b) $\dfrac{3v(v - 3)}{(v + 3)(v - 3)} = \dfrac{3v}{v + 3}\text{.}$ Undefined when $v = 3$ or $v = -3$ .

(a) $\dfrac{x + 4}{3}$ , $x \neq 4$ (b) $\dfrac{3v}{v + 3}$ , $v \neq \pm 3$

Question 3

Simplify (a) $\dfrac{x^2 + 8x + 16}{x^2 - 16}$ (b) $\dfrac{4a^2 - 4a + 1}{2a^2 - a}$

ShowHide solution

(a) $x^2 + 8x + 16 = {(x + 4)}^2\text{.}$ $\dfrac{(x + 4)(x + 4)}{(x + 4)(x - 4)} = \dfrac{x + 4}{x - 4}\text{.}$

(b) $4a^2 - 4a + 1 = {(2a - 1)}^2\text{,}$ since $(2a)^2 = 4a^2$ , $1^2 = 1$ and $2 \times 2a \times 1 = 4a$ . $2a^2 - a = a{(2a - 1)}$ .

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

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

Question 4

Simplify, stating the values for which each original fraction is undefined. (a) $\dfrac{x^2 + 7x + 12}{x^2 + x - 6}$ (b) $\dfrac{x^2 - x - 12}{x^2 - 8x + 16}$

ShowHide solution

(a) $\dfrac{(x + 3)(x + 4)}{(x + 3)(x - 2)} = \dfrac{x + 4}{x - 2}\text{.}$ The original denominator is $0$ when $x = -3$ or $x = 2$ .

(b) $\dfrac{(x - 4)(x + 3)}{(x - 4)(x - 4)} = \dfrac{x + 3}{x - 4}\text{.}$ Undefined when $x = 4$ .

Check (a) with $x = 1$ : $\dfrac{20}{-4} = -5$ , and $\dfrac{5}{-1} = -5$ .

(a) $\dfrac{x + 4}{x - 2}$ , $x \neq -3$ , $x \neq 2$ (b) $\dfrac{x + 3}{x - 4}$ , $x \neq 4$

Question 5

Simplify (a) $\dfrac{3x^2 - 5x - 2}{x^2 - 4}$ (b) $\dfrac{2x^2 + 7x + 3}{4x^2 + 2x}$

ShowHide solution

(a) $ac = -6$ , sum $-5$ : $-6$ and $1$ . $3x^2 - 6x + x - 2 = 3x{(x - 2)} + {(x - 2)} = {(3x + 1)}{(x - 2)}\text{.}$

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

(b) $ac = 6$ , sum $7$ : $6$ and $1$ . $2x^2 + 6x + x + 3 = 2x{(x + 3)} + {(x + 3)} = {(2x + 1)}{(x + 3)}\text{.}$ $4x^2 + 2x = 2x{(2x + 1)}\text{.}$

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

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

Question 6

Simplify (a) $\dfrac{a^2 - b^2}{a^2 + ab}$ (b) $\dfrac{x^2 + xy - 2y^2}{x^2 - y^2}$

ShowHide solution

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

(b) Product $-2$ , sum $1$ : $2$ and $-1$ . $x^2 + xy - 2y^2 = {(x + 2y)}{(x - y)}\text{.}$

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

(a) $\dfrac{a - b}{a}$ (b) $\dfrac{x + 2y}{x + y}$

Question 7

Simplify $\dfrac{px + py - 3x - 3y}{p^2 - 9}\text{.}$

ShowHide solution

Group: $p{(x + y)} - 3{(x + y)} = {(p - 3)}{(x + y)}\text{.}$ And $p^2 - 9 = {(p + 3)}{(p - 3)}\text{.}$

$\dfrac{(p - 3)(x + y)}{(p + 3)(p - 3)} = \dfrac{x + y}{p + 3}\text{.}$

$\dfrac{x + y}{p + 3}$

Question 8

Simplify (a) $\dfrac{5 - x}{x^2 - 25}$ (b) $\dfrac{6 - 2y}{y^2 - 3y}$ (c) $\dfrac{9 - 6x}{4x^2 - 12x + 9}$

ShowHide solution

(a) $5 - x = -{(x - 5)}$ : $\dfrac{-(x - 5)}{(x + 5)(x - 5)} = -\dfrac{1}{x + 5}\text{.}$

(b) $6 - 2y = 2{(3 - y)} = -2{(y - 3)}\text{:}$ $\dfrac{-2(y - 3)}{y(y - 3)} = -\dfrac{2}{y}\text{.}$

(c) $9 - 6x = -3{(2x - 3)}\text{,}$ and $4x^2 - 12x + 9 = {(2x - 3)}^2\text{:}$ $\dfrac{-3(2x - 3)}{(2x - 3)(2x - 3)} = -\dfrac{3}{2x - 3}\text{.}$

Check (c) with $x = 2$ : $\dfrac{9 - 12}{16 - 24 + 9} = \dfrac{-3}{1} = -3\text{,}$ and $-\dfrac{3}{1} = -3$ .

(a) $-\dfrac{1}{x + 5}$ (b) $-\dfrac{2}{y}$ (c) $-\dfrac{3}{2x - 3}$

Question 9

Which of these can be simplified? Simplify those that can. (a) $\dfrac{x + 6}{x + 3}$ (b) $\dfrac{x^2 + 9}{x + 3}$ (c) $\dfrac{x^2 + 3x}{x + 3}$ (d) $\dfrac{x^2 - 9}{x - 3}$

ShowHide solution

(a) No common factor ( $x + 6$ and $x + 3$ do not factorise). Cannot be simplified.

(b) $x^2 + 9$ is a sum of two squares and does not factorise. Cannot be simplified.

(d) $\dfrac{(x + 3)(x - 3)}{x - 3} = x + 3\text{,}$ for $x \neq 3$ .

Question 10

Find and correct the error in each. (a) Wei Ling: $\dfrac{x^2 - 25}{x + 5} = x + 5\text{.}$ (b) Siva: $\dfrac{3x^2 - 12}{6x - 12} = \dfrac{3(x^2 - 4)}{6(x - 2)} = \dfrac{x^2 - 4}{2(x - 2)}\text{,}$ and he stops.

ShowHide solution

(a) Her answer would need $x^2 - 25 = {(x + 5)}^2$ , but ${(x + 5)}^2 = x^2 + 10x + 25\text{.}$ Test $x = 1$ : $\dfrac{1 - 25}{1 + 5} = \dfrac{-24}{6} = -4\text{,}$ but $x + 5 = 6$ . Correct: $x^2 - 25 = {(x + 5)}{(x - 5)}\text{,}$ so $\dfrac{(x + 5)(x - 5)}{x + 5} = x - 5\text{,}$ for $x \neq -5$ . With $x = 1$ : $1 - 5 = -4$ .

(b) Siva's steps are right, but he has not finished: $x^2 - 4 = {(x + 2)}{(x - 2)}\text{,}$ so $\dfrac{(x + 2)(x - 2)}{2(x - 2)} = \dfrac{x + 2}{2}\text{.}$

(a) $x - 5$ (b) $\dfrac{x + 2}{2}$

Question 11

A rectangle has an area of ${(x^2 + 5x + 6)}$ cm² and a breadth of ${(x + 2)}$ cm. (a) Find its length in terms of $x$ . (b) Find the length when $x = 4$ , and check your answer from the area.

ShowHide solution

(a) Length $= \dfrac{\text{area}}{\text{breadth}} = \dfrac{x^2 + 5x + 6}{x + 2} = \dfrac{(x + 2)(x + 3)}{x + 2} = x + 3\text{.}$

(b) $4 + 3 = 7$ . Check: the area is $16 + 20 + 6 = 42$ cm² and the breadth is $6$ cm, and $42 \div 6 = 7$ .

(a) ${(x + 3)}$ cm (b) $7$ cm

Question 12

A bag of ${(n^2 + 3n)}$ sweets is shared equally among ${(n + 3)}$ children. (a) How many sweets does each child get, in terms of $n$ ? (b) There are $15$ children. How many sweets does each child get, and how many sweets were in the bag?

ShowHide solution

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

(b) $n + 3 = 15$ , so $n = 12$ : each child gets $12$ sweets. The bag held $12^2 + 3 \times 12 = 144 + 36 = 180$ sweets, and $180 \div 15 = 12$ .

(a) $n$ (b) $12$ sweets each; $180$ sweets in the bag

Key Points

Factorise the numerator and the denominator completely, then cancel the common factors.
A whole bracket is one factor: it cancels only with the same bracket, or with its reverse, leaving $-1$ .
Use every method of chapter 3: common factor first, then an identity, a quadratic or grouping.
$b - a = -{(a - b)}$ , so $\dfrac{a - b}{b - a} = -1$ .
If no factor is common after factorising completely, the fraction is already in its simplest form.
The values that make the original denominator zero stay excluded after cancelling.

Simplifying Algebraic Fractions by Factorising

Before you start

Factorise First, Then Cancel

The ways to factorise (chapter 3)

Simplifying by factorising

Example 1: A Common Factor in the Numerator or the Denominator

Example 2: The Same Bracket Top and Bottom

Using the Identities

Example 3: Difference of Two Squares

Example 4: A Perfect Square

Quadratic Expressions Top and Bottom

Example 5: Two Quadratics

Example 6: A Quadratic with a≠1a \neq 1a≠1

Example 7: Two Letters

Factorising by Grouping

Example 8: Four Terms on Top

Brackets That Are Negatives of Each Other

Reversing a bracket

Example 9: Using b−a=−(a−b)b - a = -{(a - b)}b−a=−(a−b)

When Nothing Cancels, and Finishing the Job

Example 10: Simplify If Possible

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

Example 6: A Quadratic with $a \neq 1$

Example 9: Using $b - a = -{(a - b)}$