Factorising by Grouping

How to factorise four terms such as $ax + bx + kay + kby$ into two brackets: taking out a bracket as a common factor, choosing the pairs, and dealing with minus signs.

Before you start

In Secondary 1 you factorised by taking out the highest common factor (HCF) of the terms: $6x + 9 = 3(2x + 3)$ and $5ab - 10a = 5a(b - 2)\text{.}$

In this lesson the common factor can be a whole bracket.

A Bracket as a Common Factor

In $3(x + 2) + y(x + 2)$ both terms contain the bracket $(x + 2)$ . Treat it like a single letter $B$ : $3B + yB = B(3 + y)$ . So $3(x + 2) + y(x + 2) = (x + 2)(3 + y)\text{.}$

Example 1: Taking out a common bracket

Factorise (a) $a(c + d) + b(c + d)$ ; (b) $2x(y - 5) - 3(y - 5)\text{;}$ (c) $c(a + b) + d(b + a)$ .

Solution

(a) The common factor is $(c + d)$ : $(c + d)(a + b)$ , which we may write $(a + b)(c + d)$ .

(b) The common factor is $(y - 5)$ . What is left is $2x$ from the first term and $-3$ from the second: $(y - 5)(2x - 3)$ .

Check: $(y - 5)(2x - 3) = 2xy - 3y - 10x + 15\text{,}$ and $2x(y - 5) - 3(y - 5) = 2xy - 10x - 3y + 15\text{.}$ ✓

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

Factorising by Grouping

When you expand $(a + b)(c + d)$ , the step in the middle is $a(c + d) + b(c + d)$ :

$(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd$

Factorising runs the same steps backwards. This is called factorisation by grouping, because the four terms are first put into two groups.

Factorisation by grouping

1. Group the four terms into two pairs so that the two terms of each pair have a common factor.

2. Take out the common factor of each pair. The two brackets left must be the same.

3. Take out that common bracket.

$ac + ad + bc + bd = (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d)$

Check the answer by expanding it.

Example 2: The first two and the last two

Factorise $xy + 3y + 2x + 6$ .

Solution

Group: $(xy + 3y) + (2x + 6)$

Take out each common factor: $y(x + 3) + 2(x + 3)$

Take out the common bracket: $(x + 3)(y + 2)$

Check: $(x + 3)(y + 2) = xy + 2x + 3y + 6\text{.}$ ✓

$(x + 3)(y + 2)$

Example 3: The form $ax + bx + kay + kby$

Factorise (a) $2ax + 2bx + 3ay + 3by\text{;}$ (b) $3x + 6y + 2xz + 4yz$ .

Solution

(a) $(2ax + 2bx) + (3ay + 3by) = 2x(a + b) + 3y(a + b) = (a + b)(2x + 3y)$

Check: $(a + b)(2x + 3y) = 2ax + 3ay + 2bx + 3by\text{.}$ ✓

(b) $(3x + 6y) + (2xz + 4yz) = 3(x + 2y) + 2z(x + 2y) = (x + 2y)(3 + 2z)$

(a) $(a + b)(2x + 3y)$ ; (b) $(x + 2y)(3 + 2z)$

The Frame View

The multiplication frame shows the same idea. Place the four terms in the four cells so that each row has a common factor and each column has a common factor. The common factor of each row goes on its left, and the common factor of each column goes on top. The headers are then the two brackets.

Example 4: An arrangement that works, and one that does not

Use a frame to factorise $3x + 6y + 2xz + 4yz$ . Then explain why the arrangement with $3x$ and $4yz$ in the top row does not work.

Solution

Put $3x$ and $2xz$ in the top row (common factor $x$ ), and $6y$ and $4yz$ in the bottom row (common factor $2y$ ). Then the first column $3x$ , $6y$ has common factor $3$ , and the second column $2xz$ , $4yz$ has common factor $2z$ :

×	$3$	$+2z$
$x$	$3x$	$+2xz$
$+2y$	$+6y$	$+4yz$

$3x + 2xz + 6y + 4yz = (x + 2y)(3 + 2z)$

Each cell is its row times its column, so the frame is right.

Now put $3x$ and $4yz$ in the top row:

×
	$3x$	$+4yz$
	$+6y$	$+2xz$

$3x$ and $4yz$ have no common factor except $1$ , so there is nothing useful to write on the left of the top row. This arrangement does not work. Each row and each column must have a common factor; here the top row does not.

$3x + 6y + 2xz + 4yz = (x + 2y)(3 + 2z)$

When the First Pairing Fails

The terms are not always written in a helpful order. If the first two terms have no common factor, do not give up: choose different pairs. The order in which the terms are written does not matter: choose pairs that each share a factor and leave the same bracket.

Example 5: Re-pairing

Factorise (a) $ac + bd + ad + bc$ ; (b) $xy + 12 + 3x + 4y$ .

Solution

(a) $ac$ and $bd$ have no common factor, so the pair $(ac + bd)$ is no use. Pair $ac$ with $ad$ (common factor $a$ ) and $bc$ with $bd$ (common factor $b$ ):

$(ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d)$

(b) $xy$ and $12$ have no common factor. Pair $xy$ with $3x$ and $4y$ with $12$ :

$(xy + 3x) + (4y + 12) = x(y + 3) + 4(y + 3) = (y + 3)(x + 4)$

Check: $(y + 3)(x + 4) = xy + 4y + 3x + 12\text{.}$ ✓

(a) $(a + b)(c + d)$ ; (b) $(y + 3)(x + 4)$

Minus Signs

When the second pair begins with a minus sign, take out a negative common factor, so that the bracket left matches the first one. Taking out $-b$ from $-bx + by$ gives $-b(x - y)$ : the sign inside the bracket changes.

Example 6: A negative factor from the second pair

Factorise $ax - ay - bx + by$ .

Solution

$(ax - ay) + (-bx + by)$

$= a(x - y) - b(x - y)\text{,}$ since $-b \times x = -bx$ and $-b \times (-y) = +by$ .

$= (x - y)(a - b)$

Check: $(x - y)(a - b) = ax - bx - ay + by\text{.}$ ✓

$(a - b)(x - y)$

Example 7: A factor of $-1$

Factorise $xy - 2x - y + 2$ .

Solution

$(xy - 2x) + (-y + 2)$

$= x(y - 2) - 1(y - 2)\text{.}$ The second pair has only the common factor $-1$ ; write it so that the bracket matches.

$= (y - 2)(x - 1)$

Check: $(x - 1)(y - 2) = xy - 2x - y + 2\text{.}$ ✓

$(x - 1)(y - 2)$

A Common Factor First, and Terms with Squares

Example 8: Take out the common factor of all four terms first

Factorise $6ax + 6bx + 12ay + 12by\text{.}$

Solution

All four terms have the common factor $6$ : $6(ax + bx + 2ay + 2by)\text{.}$

Group inside the bracket: $6[x(a + b) + 2y(a + b)] = 6(a + b)(x + 2y)\text{.}$

$6(a + b)(x + 2y)$

Example 9: Terms with squares

Factorise (a) $x^2 + 3x + xy + 3y$ ; (b) $x^3 + 2x^2 + 3x + 6$ .

Solution

(a) $(x^2 + 3x) + (xy + 3y) = x(x + 3) + y(x + 3) = (x + 3)(x + y)$

(b) $(x^3 + 2x^2) + (3x + 6) = x^2(x + 2) + 3(x + 2) = (x + 2)(x^2 + 3)\text{,}$ since $x^3 = x^2 \times x$ .

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

Grouping to Calculate

Example 10: A quick sum

Without a calculator, find $37 \times 13 + 37 \times 7 + 63 \times 13 + 63 \times 7\text{.}$

Solution

Group the first two and the last two: $37(13 + 7) + 63(13 + 7)$

$= (37 + 63)(13 + 7) = 100 \times 20 = 2000$

$2000$

Common Mistakes

Stopping half way: $y(x + 3) + 2(x + 3)$ is not factorised. It is still a sum; take out $(x + 3)$ .
Pairing terms with no common factor, such as $(ac + bd)$ . Choose pairs that share a factor.
Not changing the sign inside the bracket: $-bx + by = -b(x - y)\text{,}$ not $-b(x + y)$ .
Forgetting the $1$ : $x(y - 2) - (y - 2) = (y - 2)(x - 1)\text{,}$ not $(y - 2)x$ .
Leaving a common factor of all four terms inside: $6ax + 6bx + 12ay + 12by = 6(a + b)(x + 2y)\text{.}$

Practice

Try each question before you open its solution. Check each answer by expanding it.

Question 1

Factorise (a) $5(x - 4) + y(x - 4)$ ; (b) $2a(b + 3) - (b + 3)$ ; (c) $p(q - r) + 2(r - q)$ .

ShowHide solution

(a) $(x - 4)(5 + y)$

(b) $-(b + 3) = -1(b + 3)\text{:}$ $(b + 3)(2a - 1)$

(a) $(x - 4)(5 + y)$ ; (b) $(b + 3)(2a - 1)$ ; (c) $(q - r)(p - 2)$

Question 2

Factorise (a) $xy + 5x + 4y + 20$ ; (b) $6mn + 8m + 9n + 12$ .

ShowHide solution

(a) $x(y + 5) + 4(y + 5) = (y + 5)(x + 4)$

(b) $2m(3n + 4) + 3(3n + 4) = (3n + 4)(2m + 3)$

(a) $(x + 4)(y + 5)$ ; (b) $(2m + 3)(3n + 4)$

Question 3

Factorise (a) $5ax + 5bx + 2ay + 2by\text{;}$ (b) $ap + bp + 4aq + 4bq$ .

ShowHide solution

(a) $5x(a + b) + 2y(a + b) = (a + b)(5x + 2y)$

(b) $p(a + b) + 4q(a + b) = (a + b)(p + 4q)$

(a) $(a + b)(5x + 2y)$ ; (b) $(a + b)(p + 4q)$

Question 4

Factorise $2a + 4b + 3ac + 6bc$ using a multiplication frame.

ShowHide solution

Top row $2a$ and $3ac$ (common factor $a$ ); bottom row $4b$ and $6bc$ (common factor $2b$ ). Columns: $2a$ , $4b$ have common factor $2$ ; $3ac$ , $6bc$ have common factor $3c$ .

×	$2$	$+3c$
$a$	$2a$	$+3ac$
$+2b$	$+4b$	$+6bc$

$2a + 3ac + 4b + 6bc = (a + 2b)(2 + 3c)$

$(a + 2b)(2 + 3c)$

Question 5

Factorise (a) $3x + 2y + 6 + xy$ ; (b) $pq + 6 + 2p + 3q$ .

ShowHide solution

(a) $3x$ and $2y$ share no factor. Pair $xy$ with $3x$ , and $2y$ with $6$ : $x(y + 3) + 2(y + 3) = (y + 3)(x + 2)\text{.}$

(b) Pair $pq$ with $2p$ , and $3q$ with $6$ : $p(q + 2) + 3(q + 2) = (q + 2)(p + 3)\text{.}$

(a) $(x + 2)(y + 3)$ ; (b) $(p + 3)(q + 2)$

Question 6

Factorise (a) $px - py - qx + qy$ ; (b) $ab - 3a - 2b + 6$ ; (c) $xy - x - y + 1$ .

ShowHide solution

(a) $p(x - y) - q(x - y) = (x - y)(p - q)$

(b) $a(b - 3) - 2(b - 3) = (b - 3)(a - 2)\text{,}$ since $-2 \times (-3) = +6$ .

(a) $(p - q)(x - y)$ ; (b) $(a - 2)(b - 3)$ ; (c) $(x - 1)(y - 1)$

Question 7

Factorise $2xy - 6x - y + 3$ .

ShowHide solution

$2x(y - 3) - 1(y - 3) = (y - 3)(2x - 1)$

Check: $(2x - 1)(y - 3) = 2xy - 6x - y + 3\text{.}$ ✓

$(2x - 1)(y - 3)$

Question 8

Factorise (a) $4ax + 4ay + 8bx + 8by\text{;}$ (b) $3pr - 3ps + 6qr - 6qs\text{.}$

ShowHide solution

(a) $4(ax + ay + 2bx + 2by) = 4[a(x + y) + 2b(x + y)] = 4(x + y)(a + 2b)$

(b) $3(pr - ps + 2qr - 2qs) = 3[p(r - s) + 2q(r - s)] = 3(r - s)(p + 2q)$

(a) $4(a + 2b)(x + y)$ ; (b) $3(p + 2q)(r - s)$

Question 9

Factorise (a) $a^2 + 2a + ab + 2b$ ; (b) $x^3 - 4x^2 + 2x - 8$ .

ShowHide solution

(a) $a(a + 2) + b(a + 2) = (a + 2)(a + b)$

(b) $x^2(x - 4) + 2(x - 4) = (x - 4)(x^2 + 2)$

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

Question 10

Without a calculator, find $15 \times 23 + 15 \times 17 + 85 \times 23 + 85 \times 17\text{.}$

ShowHide solution

$15(23 + 17) + 85(23 + 17) = (15 + 85)(23 + 17) = 100 \times 40 = 4000$

$4000$

Question 11

A rectangle has an area of $(xy + 4x + 3y + 12)$ cm². Its length and breadth are expressions of the form $(x + p)$ cm and $(y + q)$ cm. Find them, and check with $x = 2$ and $y = 5$ .

ShowHide solution

$xy + 4x + 3y + 12 = x(y + 4) + 3(y + 4) = (x + 3)(y + 4)$

With $x = 2$ , $y = 5$ : the area is $10 + 8 + 15 + 12 = 45\text{,}$ and $(2 + 3)(5 + 4) = 5 \times 9 = 45\text{.}$ ✓

The sides are $(x + 3)$ cm and $(y + 4)$ cm.

Question 12

Mei wrote $3px - 3py - 2qx + 2qy = 3p(x - y) - 2q(x + y)$ and said it cannot be factorised further. Find her mistake and finish the factorisation.

ShowHide solution

$-2q(x + y) = -2qx - 2qy\text{,}$ but the last two terms are $-2qx + 2qy$ . Taking out $-2q$ changes the sign inside: $-2qx + 2qy = -2q(x - y)\text{.}$

So $3p(x - y) - 2q(x - y) = (x - y)(3p - 2q)\text{.}$

Check: $(x - y)(3p - 2q) = 3px - 2qx - 3py + 2qy\text{.}$ ✓

$(x - y)(3p - 2q)$

Key Points

A bracket can be a common factor: $a(c + d) + b(c + d) = (a + b)(c + d)\text{.}$
Factorise four terms by grouping them into two pairs, each with a common factor; the two brackets left must match.
If the first pairing has no common factor, pair the terms differently.
Take out a negative factor from a pair that starts with a minus: $-bx + by = -b(x - y)\text{.}$
Take out a common factor of all four terms first.
Check by expanding.

Factorising by Grouping

Before you start

A Bracket as a Common Factor

Example 1: Taking out a common bracket

Factorising by Grouping

Factorisation by grouping

Example 2: The first two and the last two

Example 3: The form ax+bx+kay+kbyax + bx + kay + kbyax+bx+kay+kby

The Frame View

Example 4: An arrangement that works, and one that does not

When the First Pairing Fails

Example 5: Re-pairing

Minus Signs

Example 6: A negative factor from the second pair

Example 7: A factor of −1-1−1

A Common Factor First, and Terms with Squares

Example 8: Take out the common factor of all four terms first

Example 9: Terms with squares

Grouping to Calculate

Example 10: A quick sum

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 3: The form $ax + bx + kay + kby$

Example 7: A factor of $-1$