Factorising Completely

How to factorise an expression completely: take out the common factor first, then use an identity, grouping or a quadratic, and check that nothing is left to factorise. With a mixed set for choosing the method.

Before you start

You now have four ways to factorise: taking out a common factor (Secondary 1), the identities (lesson 3.7), grouping (lesson 3.8), and quadratics in two brackets (lessons 3.9 to 3.11).

Many expressions need two of them, one after the other. This lesson is about doing them in the right order and finishing the job.

What "Completely" Means

Factorise completely

An expression is factorised completely when no bracket has a common factor left inside it, and no bracket can still be factorised.

1. Take out the highest common factor (HCF) of all the terms. If the first term is negative, you may take out a negative factor.

2. Count the terms left in the bracket. Two terms: is it a difference of two squares? Three terms: a perfect square, or two brackets with product $ac$ and sum $b$ ? Four terms: grouping.

3. Look at every bracket again, and check the answer by expanding it.

For example, $2x^2 - 8 = 2(x^2 - 4)$ is factorised, but not completely: $x^2 - 4$ is a difference of two squares. Completely, $2x^2 - 8 = 2(x + 2)(x - 2)\text{.}$

Common Factors with Powers

The HCF of terms with letters is the HCF of the numbers times each letter that is in every term, to the lowest power it has. For $6x^2$ and $15x$ : the HCF of $6$ and $15$ is $3$ , and $x$ is in both, at least once, so the HCF is $3x$ .

Example 1: Letters and powers

Factorise (a) $-a^3bc - a^2b^2$ ; (b) $c^2d^3 + c^3d^2 - c^2d^2$ .

Solution

(a) The letters common to both terms: $a^2$ (lowest power of $a$ ) and $b$ . The first term is negative, so take out $-a^2b$ : $-a^3bc \div (-a^2b) = ac$ and $-a^2b^2 \div (-a^2b) = b\text{.}$ So $-a^2b(ac + b)$ .

(b) HCF $c^2d^2$ : $c^2d^3 \div c^2d^2 = d$ , $c^3d^2 \div c^2d^2 = c$ , and $c^2d^2 \div c^2d^2 = 1$ . So $c^2d^2(d + c - 1)$ . The $1$ must not be left out.

(a) $-a^2b(ac + b)$ ; (b) $c^2d^2(d + c - 1)$

Example 2: A negative common factor

Factorise $-10ax + 25ay$ .

Solution

The HCF of the two terms is $5a$ . Take out $5a$ : $5a(-2x + 5y) = 5a(5y - 2x)\text{.}$ Or take out $-5a$ : $-5a(2x - 5y)$ . Both are right; they expand to the same expression.

$5a(5y - 2x)$ , or $-5a(2x - 5y)$

A Common Factor, Then an Identity

Example 3: Then a difference of two squares

Factorise completely (a) $2x^2 - 18$ ; (b) $3x^3 - 12x$ .

Solution

(a) $2x^2 - 18 = 2(x^2 - 9)\text{.}$ Two terms, both squares, a minus between: $2(x + 3)(x - 3)$ .

(b) The HCF is $3x$ : $3x^3 - 12x = 3x(x^2 - 4) = 3x(x + 2)(x - 2)\text{.}$

Neither $2x^2 - 18$ nor $3x^3 - 12x$ is itself a difference of two squares ( $2$ , $18$ , $3$ and $12$ are not square numbers); the identity appears only after the common factor is taken out.

(a) $2(x + 3)(x - 3)$ ; (b) $3x(x + 2)(x - 2)$

Example 4: Then a perfect square

Factorise completely (a) $2x^2 + 12x + 18$ ; (b) $3a^2 - 12ab + 12b^2$ .

Solution

(a) $2(x^2 + 6x + 9)$ . $9 = 3^2$ and $2 \times x \times 3 = 6x$ : $2(x + 3)^2$ .

(b) $3(a^2 - 4ab + 4b^2)$ . $4b^2 = (2b)^2$ and $2 \times a \times 2b = 4ab$ , with a minus: $3(a - 2b)^2$ .

(a) $2(x + 3)^2$ ; (b) $3(a - 2b)^2$

A Common Factor, Then a Quadratic

Example 5: A number or a letter first

Factorise completely (a) $3x^2 - 3x - 36$ ; (b) $x^3 + 5x^2 + 6x$ .

Solution

(a) $3(x^2 - x - 12)$ . Product $-12$ , sum $-1$ : $-4$ and $3$ . So $3(x - 4)(x + 3)$ .

(b) Every term has $x$ : $x(x^2 + 5x + 6)$ . Product $6$ , sum $5$ : $2$ and $3$ . So $x(x + 2)(x + 3)$ .

Check (b): $x(x + 2)(x + 3) = x(x^2 + 5x + 6) = x^3 + 5x^2 + 6x\text{.}$ ✓

(a) $3(x - 4)(x + 3)$ ; (b) $x(x + 2)(x + 3)$

Example 6: A negative common factor first

Factorise completely (a) $-x^2 + 5x - 6$ ; (b) $-2x^2 + 8$ .

Solution

(a) Take out $-1$ : $-(x^2 - 5x + 6) = -(x - 2)(x - 3)\text{.}$

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

Check (a): $-(x - 2)(x - 3) = -(x^2 - 5x + 6) = -x^2 + 5x - 6\text{.}$ ✓

(a) $-(x - 2)(x - 3)$ ; (b) $-2(x + 2)(x - 2)$

A Bracket as the Common Factor

Example 7: A common bracket

Factorise (a) $3x(x - 2) - 5(x - 2)\text{;}$ (b) $(x + 1)^2 - 3(x + 1)$ .

Solution

(a) The common factor is $(x - 2)$ : $(x - 2)(3x - 5)$ .

(b) $(x + 1)^2 = (x + 1)(x + 1)\text{,}$ so $(x + 1)$ is common: $(x + 1)\big((x + 1) - 3\big) = (x + 1)(x - 2)\text{.}$

Check (b): $(x + 1)(x - 2) = x^2 - x - 2\text{,}$ and $(x + 1)^2 - 3(x + 1) = x^2 + 2x + 1 - 3x - 3 = x^2 - x - 2\text{.}$ ✓

(a) $(x - 2)(3x - 5)$ ; (b) $(x + 1)(x - 2)$

Finishing a Factorisation

Example 8: Not finished yet

Each of these is correct but not factorised completely. Finish it. (a) $(3x + 3)(2x - 5)$ ; (b) $(4x - 6)(x + 1)$ ; (c) $2(x^2 - 4)$

Solution

(a) $3x + 3 = 3(x + 1)$ , so $(3x + 3)(2x - 5) = 3(x + 1)(2x - 5)\text{.}$

(b) $4x - 6 = 2(2x - 3)$ , so $2(2x - 3)(x + 1)$ .

(a) $3(x + 1)(2x - 5)$ ; (b) $2(2x - 3)(x + 1)$ ; (c) $2(x + 2)(x - 2)$

Choosing the Method

After the common factor, the number of terms left points to the method:

Terms left	Look for	Example
two	a difference of two squares (a sum of two squares does not factorise)	$x^2 - 25 = (x + 5)(x - 5)$
three	a perfect square, or two numbers with product $ac$ and sum $b$	$x^2 - 14x + 49 = (x - 7)^2$
four	grouping in pairs	$ax + ay - 3x - 3y = (x + y)(a - 3)$

Example 9: A mixed set

Factorise completely, if possible. (a) $5x^2 - 20$ ; (b) $ax + ay - 3x - 3y$ ; (c) $x^2 - 14x + 49$ ; (d) $2x^2 + 5x - 3$ ; (e) $x^2 + 4$ ; (f) $3x^2 + 6x + 9$

Solution

(a) Common factor $5$ : $5(x^2 - 4)$ ; then two squares: $5(x + 2)(x - 2)$ .

(b) No common factor of all four; group: $a(x + y) - 3(x + y) = (x + y)(a - 3)\text{.}$

(d) No common factor. $ac = -6$ , sum $5$ : $6$ and $-1$ . $2x^2 + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3)\text{.}$

(e) No common factor, and a sum of two squares: it cannot be factorised.

(f) Common factor $3$ : $3(x^2 + 2x + 3)$ . For $x^2 + 2x + 3$ we need product $3$ and sum $2$ ; the pairs $1, 3$ and $-1, -3$ have sums $4$ and $-4$ . So $3(x^2 + 2x + 3)$ is complete.

(a) $5(x + 2)(x - 2)$ ; (b) $(x + y)(a - 3)$ ; (c) $(x - 7)^2$ ; (d) $(2x - 1)(x + 3)$ ; (e) cannot be factorised; (f) $3(x^2 + 2x + 3)$

Factorising to Calculate

Example 10: A common factor and a difference of squares

Without a calculator, find $2 \times 53^2 - 2 \times 47^2$ .

Solution

$2 \times 53^2 - 2 \times 47^2 = 2(53^2 - 47^2) = 2(53 + 47)(53 - 47)$

$= 2 \times 100 \times 6 = 1200$

$1200$

Common Mistakes

Stopping after the common factor: $2x^2 - 8 = 2(x^2 - 4)$ is not complete; it is $2(x + 2)(x - 2)$ .
Leaving a common factor in a bracket: $(3x + 3)(2x - 5)$ should be $3(x + 1)(2x - 5)$ .
Losing the $1$ when a whole term is taken out: $c^2d^3 + c^3d^2 - c^2d^2 = c^2d^2(d + c - 1)\text{.}$
Losing the negative factor: $-x^2 + 5x - 6 = -(x - 2)(x - 3)\text{,}$ not $(x - 2)(x - 3)$ .
Trying to factorise a sum of two squares such as $x^2 + 4$ .

Practice

Factorise each expression completely. Check each answer by expanding it.

Question 1

(a) $8x^2 - 12x$ ; (b) $6a^2b + 9ab^2$ ; (c) $x^3y - x^2y^2 + x^2y$

ShowHide solution

(a) HCF $4x$ : $4x(2x - 3)$ .

(b) HCF $3ab$ : $3ab(2a + 3b)$ .

(a) $4x(2x - 3)$ ; (b) $3ab(2a + 3b)$ ; (c) $x^2y(x - y + 1)$

Question 2

$-4x^2 + 10x$

ShowHide solution

The first term is negative, so take out $-2x$ : $-2x(2x - 5)$ .

$-2x(2x - 5)$

Question 3

(a) $3x^2 - 75$ ; (b) $8y^2 - 2$ ; (c) $x^3 - 9x$

ShowHide solution

(a) $3(x^2 - 25) = 3(x + 5)(x - 5)$

(b) $2(4y^2 - 1) = 2(2y + 1)(2y - 1)$

(a) $3(x + 5)(x - 5)$ ; (b) $2(2y + 1)(2y - 1)$ ; (c) $x(x + 3)(x - 3)$

Question 4

(a) $5x^2 + 20x + 20$ ; (b) $2a^2 - 24a + 72$

ShowHide solution

(a) $5(x^2 + 4x + 4) = 5(x + 2)^2$

(b) $2(a^2 - 12a + 36) = 2(a - 6)^2$

(a) $5(x + 2)^2$ ; (b) $2(a - 6)^2$

Question 5

(a) $2x^2 + 2x - 24$ ; (b) $x^3 - x^2 - 6x$ ; (c) $6x^2 + 3x - 18$

ShowHide solution

(a) $2(x^2 + x - 12)$ ; product $-12$ , sum $1$ : $4$ , $-3$ . So $2(x + 4)(x - 3)$ .

(b) $x(x^2 - x - 6)$ ; product $-6$ , sum $-1$ : $-3$ , $2$ . So $x(x - 3)(x + 2)$ .

(c) $3(2x^2 + x - 6)$ ; $ac = -12$ , sum $1$ : $4$ , $-3$ . $2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2)\text{.}$ So $3(2x - 3)(x + 2)$ .

(a) $2(x + 4)(x - 3)$ ; (b) $x(x - 3)(x + 2)$ ; (c) $3(2x - 3)(x + 2)$

Question 6

(a) $-x^2 - 2x + 15$ ; (b) $-3x^2 + 27$

ShowHide solution

(a) $-(x^2 + 2x - 15)$ ; product $-15$ , sum $2$ : $5$ , $-3$ . So $-(x + 5)(x - 3)$ .

(b) $-3(x^2 - 9) = -3(x + 3)(x - 3)$

(a) $-(x + 5)(x - 3)$ ; (b) $-3(x + 3)(x - 3)$

Question 7

(a) $2x(x + 4) + 3(x + 4)\text{;}$ (b) $(x - 3)^2 - 2(x - 3)$ ; (c) $x(x - 2) + 4(2 - x)$

ShowHide solution

(a) $(x + 4)(2x + 3)$

(b) $(x - 3)\big((x - 3) - 2\big) = (x - 3)(x - 5)$

(a) $(x + 4)(2x + 3)$ ; (b) $(x - 3)(x - 5)$ ; (c) $(x - 2)(x - 4)$

Question 8

Each answer is correct but not complete. Finish it. (a) $(3x + 6)(x - 1)$ ; (b) $(2x - 4)(2x + 4)$ ; (c) $x(x^2 - 16)$

ShowHide solution

(a) $3x + 6 = 3(x + 2)$ : $3(x + 2)(x - 1)$ .

(b) Each bracket has the factor $2$ : $2(x - 2) \times 2(x + 2) = 4(x - 2)(x + 2)\text{.}$

(a) $3(x + 2)(x - 1)$ ; (b) $4(x + 2)(x - 2)$ ; (c) $x(x + 4)(x - 4)$

Question 9

(a) $3ax - 3ay + 6bx - 6by\text{;}$ (b) $x^3 + x^2 - 4x - 4$

ShowHide solution

(a) Common factor $3$ : $3(ax - ay + 2bx - 2by) = 3[a(x - y) + 2b(x - y)] = 3(x - y)(a + 2b)\text{.}$

(b) Group: $x^2(x + 1) - 4(x + 1) = (x + 1)(x^2 - 4)\text{.}$ The bracket $x^2 - 4$ still factorises: $(x + 1)(x + 2)(x - 2)\text{.}$

(a) $3(a + 2b)(x - y)$ ; (b) $(x + 1)(x + 2)(x - 2)$

Question 10

Choose the method. Factorise completely, if possible: (a) $4x^2 - 9y^2$ ; (b) $x^2 + 8x + 16$ ; (c) $3x^2 - 10x + 8$ ; (d) $2x^2 + 8$ ; (e) $pq - 3p + 2q - 6$ ; (f) $12x^2 - 27$

ShowHide solution

(a) Two squares, minus: $(2x + 3y)(2x - 3y)$ .

(b) Perfect square: $16 = 4^2$ , $2 \times x \times 4 = 8x$ : $(x + 4)^2$ .

(d) $2(x^2 + 4)$ ; $x^2 + 4$ is a sum of two squares, so this is complete.

(e) Grouping: $p(q - 3) + 2(q - 3) = (q - 3)(p + 2)\text{.}$

(f) $3(4x^2 - 9) = 3(2x + 3)(2x - 3)\text{.}$

(a) $(2x + 3y)(2x - 3y)$ ; (b) $(x + 4)^2$ ; (c) $(3x - 4)(x - 2)$ ; (d) $2(x^2 + 4)$ ; (e) $(p + 2)(q - 3)$ ; (f) $3(2x + 3)(2x - 3)$

Question 11

Without a calculator, find (a) $3 \times 17^2 - 3 \times 13^2$ ; (b) $7 \times 26^2 - 7 \times 24^2$ .

ShowHide solution

(a) $3(17 + 13)(17 - 13) = 3 \times 30 \times 4 = 360$

(b) $7(26 + 24)(26 - 24) = 7 \times 50 \times 2 = 700$

(a) $360$ ; (b) $700$

Question 12

A box is a cuboid with volume $(2x^3 + 10x^2 + 12x)$ cm³. (a) Factorise the expression completely. (b) Suggest a length, breadth and height for the box in terms of $x$ . (c) Check when $x = 1$ .

ShowHide solution

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

(b) For example, $2x$ cm, $(x + 2)$ cm and $(x + 3)$ cm.

(a) $2x(x + 2)(x + 3)$ ; (b) for example $2x$ cm by $(x + 2)$ cm by $(x + 3)$ cm

Key Points

Factorised completely: no common factor left in any bracket, and no bracket that can still be factorised.
Always take out the HCF first, including letters to their lowest power, and a negative factor if you wish.
Then count the terms: two, a difference of squares; three, a perfect square or $ac$ and $b$ ; four, grouping.
A bracket can be the common factor: $3x(x - 2) - 5(x - 2) = (x - 2)(3x - 5)\text{.}$
A sum of two squares, such as $x^2 + 4$ , cannot be split into two brackets (a common factor may still come out: $4x^2 + 36 = 4(x^2 + 9)\text{).}$
Check by expanding.