Factorising quadratic expressions

6 Factorising quadratic expressions

Quadratic expressions commonly occur in many areas of mathematics, physics and engineering. Many quadratic expressions can be written as the product of two linear factors and, in this Section, we examine how these factors can be easily found.

Key Point 18

An expression of the form

a x^{2} + b x + c a \neq 0

where $a$ , $b$ and $c$ are numbers is called a quadratic expression (in the variable $x$ ).

The numbers $b$ and $c$ may be zero but $a$ must not be zero (for, then, the quadratic reduces to a linear expression or constant). The number $a$ is called the coefficient of $x^{2}$ , $b$ is the coefficient of $x$ and $c$ is called the constant term .

6.1 Case 1

Consider the product $(x + 1) (x + 2)$ . Removing brackets yields $x^{2} + 3 x + 2$ . Conversely, we see that the factors of $x^{2} + 3 x + 2$ are $(x + 1)$ and $(x + 2)$ . However, if we were given the quadratic expression first, how would we factorise it ? The following examples show how to do this but note that not all quadratic expressions can be easily factorised.

To enable us to factorise a quadratic expression in which the coefficient of $x^{2}$ equals 1, we note the following expansion:

$(x + m) (x + n) = x^{2} + m x + n x + m n = x^{2} + (m + n) x + m n$

So, given a quadratic expression we can think of the coefficient of $x$ as $m + n$ and the constant term as $m n$ . Once the values of $m$ and $n$ have been found the factors can be easily obtained.

Example 46

Factorise $x^{2} + 4 x - 5$ .

Solution

Writing $x^{2} + 4 x - 5 = (x + m) (x + n) = x^{2} + (m + n) x + m n$ we seek numbers $m$ and $n$ such that $m + n = 4$ and $m n = - 5$ . By trial and error it is not difficult to find that $m = 5$ and $n = - 1$ (or, the other way round, $m = - 1$ and $n = 5$ ). So we can write

$x^{2} + 4 x - 5 = (x + 5) (x - 1)$

The answer can be checked easily by removing brackets.

Task!

Factorise $x^{2} + 6 x + 8$ .

As the coefficient of $x^{2}$ is 1, we can write

$x^{2} + 6 x + 8 = (x + m) (x + n) = x^{2} + (m + n) x + m n$

so that $m + n = 6$ and $m n = 8$ .

First, find suitable values for $m$ and $n$ :

$m = 4, n = 2$ or, the other way round, $m = 2, n = 4$

Finally factorise the quadratic:

$(x + 4) (x + 2)$

6.2 Case 2

When the coefficient of $x^{2}$ is not equal to 1 it may be possible to extract a numerical factor. For example, note that $3 x^{2} + 18 x + 24$ can be written as $3 (x^{2} + 6 x + 8)$ and then factorised as in the previous Task in Case 1. Sometimes no numerical factor can be found and a slightly different approach may be taken. We will demonstrate a technique which can always be used to transform the given expression into one in which the coefficient of the squared variable equals 1.

Example 47

Factorise $2 x^{2} + 5 x + 3$ .

Solution

First note the coefficient of $x^{2}$ ; in this case 2. Multiply the whole expression by this number and rearrange as follows:

$2 (2 x^{2} + 5 x + 3) = 2 (2 x^{2}) + 2 (5 x) + 2 (3) = {(2 x)}^{2} + 5 (2 x) + 6.$

We now introduce a new variable $z$ such that $z = 2 x$ Thus we can write

${(2 x)}^{2} + 5 (2 x) + 6 as z^{2} + 5 z + 6$

This can be factorised to give $(z + 3) (z + 2)$ . Returning to the original variable by replacing $z$ by $2 x$ we find

$2 (2 x^{2} + 5 x + 3) = (2 x + 3) (2 x + 2)$

A factor of 2 can be extracted from the second bracket on the right so that

$2 (2 x^{2} + 5 x + 3) = 2 (2 x + 3) (x + 1)$

so that

$2 x^{2} + 5 x + 3 = (2 x + 3) (x + 1)$

As an alternative to the technique of Example 47, experience and practice can often help us to identify factors. For example suppose we wish to factorise $3 x^{2} + 7 x + 2$ . We write

$3 x^{2} + 7 x + 2 = () ()$

In order to obtain the term $3 x^{2}$ we can place terms $3 x$ and $x$ in the brackets to give

$3 x^{2} + 7 x + 2 = (3 x + ?) (x + ?)$

In order to obtain the constant 2, we consider the factors of 2. These are 1,2 or $- 1$ , $- 2$ . By placing these factors in the brackets we can factorise the quadratic expression. Various possibilities exist: we could write $(3 x + 2) (x + 1)$ or $(3 x + 1) (x + 2)$ or $(3 x - 2) (x - 1)$ or $(3 x - 1) (x - 2)$ , only one of which is correct. By removing brackets from each in turn we look for the factorisation which produces the correct middle term, $7 x$ . The correct factorisation is found to be

$3 x^{2} + 7 x + 2 = (3 x + 1) (x + 2)$

With practice you will be able to carry out this process quite easily.

Task!

Factorise the quadratic expression $5 x^{2} - 7 x - 6$ .

Write $5 x^{2} - 7 x - 6 = () ()$

To obtain the quadratic term $5 x^{2}$ , insert $5 x$ and $x$ in the brackets:

$5 x^{2} - 7 x - 6 = (5 x + ?) (x + ?)$

Now find the factors of $- 6$ :

$3, - 2$ or $- 3, 2$ or $- 6, 1$ or $6, - 1$

Use these factors in turn to find which pair, if any, gives rise to the middle term, $- 7 x$ , and complete the factorisation:

$(5 x + 3) (x - 2)$

On occasions you will meet expressions of the form $x^{2} - y^{2}$ known as the difference of two squares .

It is easy to verify by removing brackets that this factorises as

$x^{2} - y^{2} = (x + y) (x - y)$

So, if you can learn to recognise such expressions it is an easy matter to factorise them.

Example 48

Factorise

$x^{2} - 36 z^{2}$ ,
$25 x^{2} - 9 z^{2}$ ,
$α^{2} - 1$

Solution

In each case we are required to find the difference of two squared terms.

Note that $x^{2} - 36 z^{2} = x^{2} - {(6 z)}^{2}$ . This factorises as $(x + 6 z) (x - 6 z)$ .
Here $25 x^{2} - 9 z^{2} = {(5 x)}^{2} - {(3 z)}^{2}$ . This factorises as $(5 x + 3 z) (5 x - 3 z)$ .
$α^{2} - 1 = (α + 1) (α - 1)$ .

Exercises

Factorise
1. $x^{2} + 8 x + 7$ ,
2. $x^{2} + 6 x - 7$ ,
3. $x^{2} + 7 x + 10$ ,
4. $x^{2} - 6 x + 9$ .
Factorise
1. $2 x^{2} + 3 x + 1$ ,
2. $2 x^{2} + 4 x + 2$ ,
3. $3 x^{2} - 3 x - 6$ ,
4. $5 x^{2} - 4 x - 1$ ,
5. $16 x^{2} - 1$ ,
6. $- x^{2} + 1$ ,
7. $- 2 x^{2} + x + 3$ .

Factorise

(a) $x^{2} + 9 x + 14$ ,	(b) $x^{2} + 11 x + 18$ ,	(c) $x^{2} + 7 x - 18$ ,	(d) $x^{2} + 4 x - 77$ ,
(e) $x^{2} + 2 x$ ,	(f) $3 x^{2} + x$ ,	(g) $3 x^{2} + 4 x + 1$ ,	(h) $6 x^{2} + 5 x + 1$ ,
(i) $6 x^{2} + 31 x + 35$ ,	(j) $6 x^{2} + 7 x - 5$ ,	(k) $- 3 x^{2} + 2 x + 5$ ,	(l) $x^{2} - 3 x + 2$ .

Factorise
1. $z^{2} - 144$ ,
2. $z^{2} - \frac{1}{4}$ ,
3. $s^{2} - \frac{1}{9}$

1. $(x + 7) (x + 1)$ ,
2. $(x + 7) (x - 1)$ ,
3. $(x + 2) (x + 5)$ ,
4. $(x - 3) (x - 3)$
1. $(2 x + 1) (x + 1)$ ,
2. $2 {(x + 1)}^{2}$ ,
3. $3 (x + 1) (x - 2)$ , (d) $(5 x + 1) (x - 1)$ ,
  (e) $(4 x + 1) (4 x - 1)$ , (f) $(x + 1) (1 - x)$ , (g) $(x + 1) (3 - 2 x)$

The factors are:

(a) $(7 + x) (2 + x)$ ,	(b) $(9 + x) (2 + x)$ ,	(c) $(x + 9) (x - 2)$ ,	(d) $(x + 11) (x - 7)$ ,
(e) $(x + 2) x$ ,	(f) $(3 x + 1) x$ ,	(g) $(3 x + 1) (x + 1)$ ,	(h) $(3 x + 1) (2 x + 1)$ ,
(i) $(3 x + 5) (2 x + 7)$ ,	(j) $(3 x + 5) (2 x - 1)$ ,	(k) $(5 - 3 x) (x + 1)$ ,	(l) $(2 - x) (1 - x)$ .

1. $(z + 12) (z - 12)$ ,
2. $(z + \frac{1}{2}) (z - \frac{1}{2})$ ,
3. $(s + \frac{1}{3}) (s - \frac{1}{3})$

6 Factorising quadratic expressions

6.1 Case 1

Answer

Answer

6.2 Case 2

Answer

Answer

Answer