Expanding Longer Products and Simplifying

How to expand products with more letters, a bracket of three terms, or a number in front of two brackets, and how to simplify sums and differences of products.

Before you start

In the last lesson you expanded two brackets by multiplying each term in the first by each term in the second: $(x + 4)(x - 2) = x^2 - 2x + 4x - 8 = x^2 + 2x - 8\text{.}$

The same rule expands every product in this lesson. What changes is how many products there are, and how carefully the signs must be kept.

Brackets with Different Letters

When the two brackets have no letter in common, none of the four products are like terms, so there is nothing to collect. Write the letters of each term in alphabetical order: $x \times 3c = 3cx$ .

Example 1: Four different letters

Expand (a) $(a + b)(7x + 5y)$ ; (b) $(x - 6y)(3c + 2d)$ .

Solution

(a) $a \times 7x + a \times 5y + b \times 7x + b \times 5y = 7ax + 5ay + 7bx + 5by$

(b) $x \times 3c + x \times 2d + (-6y) \times 3c + (-6y) \times 2d = 3cx + 2dx - 18cy - 12dy$

(a) $7ax + 5ay + 7bx + 5by\text{;}$ (b) $3cx + 2dx - 18cy - 12dy$

Example 2: In a frame

Expand $(8p - 3q)(2r - 5s)$ .

Solution

×	$2r$	$-5s$
$8p$	$16pr$	$-40ps$
$-3q$	$-6qr$	$+15qs$

$16pr - 40ps - 6qr + 15qs$

The four terms are all unlike, so this is the answer. Note $(-3q) \times (-5s) = +15qs\text{.}$

$(8p - 3q)(2r - 5s) = 16pr - 40ps - 6qr + 15qs$

Two Letters in Each Bracket

When both brackets use the same two letters, the products give terms in $x^2$ , $xy$ and $y^2$ . The two $xy$ terms are like terms (since $yx = xy$ ) and must be collected.

Example 3: Terms in $p^2$ , $pq$ and $q^2$

Expand (a) $(2p - q)(3p - 4q)$ ; (b) $(x + 2y)(3x - y)$ .

Solution

(a) $2p \times 3p = 6p^2$ , $2p \times (-4q) = -8pq\text{,}$ $(-q) \times 3p = -3pq$ , $(-q) \times (-4q) = +4q^2\text{.}$

$6p^2 - 8pq - 3pq + 4q^2 = 6p^2 - 11pq + 4q^2$

(b) $3x^2 - xy + 6xy - 2y^2 = 3x^2 + 5xy - 2y^2$

(a) $6p^2 - 11pq + 4q^2$ ; (b) $3x^2 + 5xy - 2y^2$

A Bracket with Three Terms

More terms, more products

Each term in the first bracket multiplies each term in the second, however many there are. Two terms times three terms give $2 \times 3 = 6$ products:

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

The frame simply gets another column.

Example 4: A frame with three columns

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

Solution

×	$x^2$	$-3x$	$+1$
$x$	$x^3$	$-3x^2$	$+x$
$+2$	$+2x^2$	$-6x$	$+2$

$x^3 - 3x^2 + x + 2x^2 - 6x + 2$

$x \times x^2 = x^3$ . Collect: $-3x^2 + 2x^2 = -x^2$ and $x - 6x = -5x$ .

$(x + 2)(x^2 - 3x + 1) = x^3 - x^2 - 5x + 2$

Example 5: Three terms in the first bracket

Expand and simplify $(x + y - 1)(x - 2)$ .

Solution

$x(x - 2) = x^2 - 2x$

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

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

Add the six terms: $x^2 - 2x + xy - 2y - x + 2\text{.}$ The like terms are $-2x$ and $-x$ .

$(x + y - 1)(x - 2) = x^2 + xy - 3x - 2y + 2$

Sometimes only one term of the answer is wanted. Then find just the products that give that term.

Example 6: One coefficient only

Find the coefficient of $x$ in the expansion of $(2x - 1)(x^2 + 3x - 4)\text{.}$

Solution

A term in $x$ comes from (a term in $x$ ) times (a number). There are two such products:

$2x \times (-4) = -8x$ and $(-1) \times 3x = -3x$ .

Together: $-8x - 3x = -11x$ .

Check by the full expansion: $2x^3 + 6x^2 - 8x - x^2 - 3x + 4 = 2x^3 + 5x^2 - 11x + 4\text{.}$ ✓

The coefficient of $x$ is $-11$ .

A Number or a Term in Front of Two Brackets

In $3(x + 1)(x - 2)$ three things are multiplied. Expand the two brackets first and keep the result in a bracket; then multiply every term by the number in front.

Example 7: A number in front

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

Solution

(a) $(x + 1)(x - 2) = x^2 - 2x + x - 2 = x^2 - x - 2$

$3(x^2 - x - 2) = 3x^2 - 3x - 6$

(b) $(x - 3)(2x + 1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$

$-2(2x^2 - 5x - 3) = -4x^2 + 10x + 6\text{.}$ Every sign changes, because $-2$ is negative.

(a) $3x^2 - 3x - 6$ ; (b) $-4x^2 + 10x + 6$

Example 8: A letter in front

Expand $x(x + 1)(x + 2)$ .

Solution

$(x + 1)(x + 2) = x^2 + 3x + 2$

$x(x^2 + 3x + 2) = x^3 + 3x^2 + 2x$

Check with $x = 2$ : $2 \times 3 \times 4 = 24$ , and $8 + 12 + 4 = 24$ . ✓

$x(x + 1)(x + 2) = x^3 + 3x^2 + 2x$

Sums and Differences of Products

A minus sign before a product

Expand the product inside a bracket first. Then remove the bracket: the minus sign changes the sign of every term of the expansion, not only the first.

$-(x - 1)(x + 4) = -(x^2 + 3x - 4) = -x^2 - 3x + 4$

Example 9: The difference of two products

Simplify $(x + 3)(x - 2) - (x - 1)(x + 4)\text{.}$

Solution

$(x + 3)(x - 2) = x^2 + x - 6$

$(x - 1)(x + 4) = x^2 + 3x - 4$

$(x^2 + x - 6) - (x^2 + 3x - 4) = x^2 + x - 6 - x^2 - 3x + 4$

$= -2x - 2$ . The $x^2$ terms cancel.

Check with $x = 1$ : $(4)(-1) - (0)(5) = -4\text{,}$ and $-2 - 2 = -4$ . ✓

$-2x - 2$

Example 10: A term times a bracket, minus two brackets

Simplify $2x(x - 5) - (x - 3)(x + 2)\text{.}$

Solution

$2x(x - 5) = 2x^2 - 10x$

$(x - 3)(x + 2) = x^2 - x - 6$

$2x^2 - 10x - (x^2 - x - 6) = 2x^2 - 10x - x^2 + x + 6$

$= x^2 - 9x + 6$

$x^2 - 9x + 6$

When everything with the letter cancels, the expression has the same value for every value of the letter. Expanding shows why.

Example 11: Always the same

Show that $(n + 1)(n + 2) - n(n + 3) = 2$ for every value of $n$ . Check it with $n = 5$ .

Solution

$(n + 1)(n + 2) = n^2 + 3n + 2$ and $n(n + 3) = n^2 + 3n$ .

$(n^2 + 3n + 2) - (n^2 + 3n) = n^2 + 3n + 2 - n^2 - 3n = 2$

With $n = 5$ : $6 \times 7 - 5 \times 8 = 42 - 40 = 2\text{.}$ ✓

The $n^2$ and $n$ terms cancel, leaving $2$ for every $n$ .

Areas

Example 12: A path round a lawn

A rectangular lawn $x$ m by $(x + 3)$ m has a path $2$ m wide all round it, as shown. Find, in its simplest form, an expression for the area of the path. Then find the area when $x = 10$ .

Solution

The path adds $2$ m at each end of each side, so the outer rectangle is $(x + 3 + 4)$ m by $(x + 4)$ m, that is $(x + 7)$ m by $(x + 4)$ m.

Area of path $=$ outer area $-$ lawn area $= (x + 4)(x + 7) - x(x + 3)$

$= (x^2 + 11x + 28) - (x^2 + 3x) = x^2 + 11x + 28 - x^2 - 3x = 8x + 28$

When $x = 10$ : $8(10) + 28 = 108$ . Check: the outer rectangle is $14$ m by $17$ m, area $238$ m²; the lawn is $10$ m by $13$ m, area $130$ m²; $238 - 130 = 108$ . ✓

The path has area $(8x + 28)$ m², which is $108$ m² when $x = 10$ .

Common Mistakes

Changing only the first sign after a minus: $-(x - 1)(x + 4)$ is $-(x^2 + 3x - 4) = -x^2 - 3x + 4\text{,}$ not $-x^2 + 3x - 4$ . Keep the expansion in a bracket until the minus has been applied to every term.
Leaving out products when a bracket has three terms: $(a + b)(c + d + e)$ has six products.
Multiplying the number in front into both brackets: $3(x + 1)(x - 2)$ is not $(3x + 3)(3x - 6)$ . The $3$ multiplies once.
Not collecting the two $xy$ terms: in $(x + 2y)(3x - y)$ , $-xy + 6xy = 5xy$ .
Leaving the answer uncollected, such as $x^2 + x - 6 - x^2 - 3x + 4\text{.}$ Simplify fully: $-2x - 2$ .

Practice

Try each question before you open its solution.

Question 1

Expand (a) $(a + b)(8x + 7y)$ ; (b) $(5a + 2)(x - 2y)$ ; (c) $(6p - 5q)(3r - 4s)$ .

ShowHide solution

(a) $8ax + 7ay + 8bx + 7by$

(b) $5a \times x = 5ax$ , $5a \times (-2y) = -10ay\text{,}$ $2 \times x = 2x$ , $2 \times (-2y) = -4y$ .

(a) $8ax + 7ay + 8bx + 7by\text{;}$ (b) $5ax - 10ay + 2x - 4y\text{;}$ (c) $18pr - 24ps - 15qr + 20qs$

Question 2

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

ShowHide solution

(a) $3x^2 + 12xy - xy - 4y^2 = 3x^2 + 11xy - 4y^2$

(b) $4a^2 - 2ab - 6ab + 3b^2 = 4a^2 - 8ab + 3b^2$

(a) $3x^2 + 11xy - 4y^2$ ; (b) $4a^2 - 8ab + 3b^2$

Question 3

Expand and simplify (a) $(x - 3)(x^2 + 2x - 5)\text{;}$ (b) $(p + 2)(q - r + 3)$ ; (c) $(2x + y + 3)(x - y)$ .

ShowHide solution

(a) $x^3 + 2x^2 - 5x - 3x^2 - 6x + 15 = x^3 - x^2 - 11x + 15$

(b) $p(q - r + 3) + 2(q - r + 3) = pq - pr + 3p + 2q - 2r + 6$

(a) $x^3 - x^2 - 11x + 15$ ; (b) $pq - pr + 3p + 2q - 2r + 6\text{;}$ (c) $2x^2 - xy - y^2 + 3x - 3y$

Question 4

Without expanding fully, find (a) the coefficient of $x^2$ and (b) the coefficient of $x$ in $(3x + 2)(x^2 - 4x + 5)\text{.}$

ShowHide solution

(a) $x^2$ terms: $3x \times (-4x) = -12x^2$ and $2 \times x^2 = 2x^2$ ; together $-10x^2$ .

(b) $x$ terms: $3x \times 5 = 15x$ and $2 \times (-4x) = -8x$ ; together $7x$ .

Check: $3x^3 - 12x^2 + 15x + 2x^2 - 8x + 10 = 3x^3 - 10x^2 + 7x + 10\text{.}$ ✓

(a) $-10$ ; (b) $7$

Question 5

Expand (a) $4(x - 2)(x + 5)$ ; (b) $-3(2x - 1)(x - 4)$ ; (c) $2x(x + 4)(x - 1)$ .

ShowHide solution

(a) $(x - 2)(x + 5) = x^2 + 3x - 10\text{;}$ times $4$ : $4x^2 + 12x - 40$ .

(b) $(2x - 1)(x - 4) = 2x^2 - 9x + 4\text{;}$ times $-3$ : $-6x^2 + 27x - 12$ .

(a) $4x^2 + 12x - 40$ ; (b) $-6x^2 + 27x - 12$ ; (c) $2x^3 + 6x^2 - 8x$

Question 6

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

ShowHide solution

(a) $(x^2 + 6x + 5) - (x^2 + 5x + 6) = x^2 + 6x + 5 - x^2 - 5x - 6 = x - 1$

(b) $(2x - 1)(x + 3) = 2x^2 + 5x - 3$ and $2(x - 2)(x + 1) = 2(x^2 - x - 2) = 2x^2 - 2x - 4\text{.}$

$(2x^2 + 5x - 3) - (2x^2 - 2x - 4) = 2x^2 + 5x - 3 - 2x^2 + 2x + 4 = 7x + 1$

(a) $x - 1$ ; (b) $7x + 1$

Question 7

Simplify (a) $3x(2x + 1) - (x + 4)(x - 4)\text{;}$ (b) $(a + 2b)(a - b) + b(3a - b)\text{.}$

ShowHide solution

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

$6x^2 + 3x - (x^2 - 16) = 6x^2 + 3x - x^2 + 16 = 5x^2 + 3x + 16$

(b) $a^2 - ab + 2ab - 2b^2 + 3ab - b^2 = a^2 + 4ab - 3b^2$

(a) $5x^2 + 3x + 16$ ; (b) $a^2 + 4ab - 3b^2$

Question 8

Expand (a) $(x^2 + 3)(x - 2)$ ; (b) $(ab + 1)(a - b)$ .

ShowHide solution

(a) $x^2 \times x = x^3$ , $x^2 \times (-2) = -2x^2$ , $3 \times x = 3x$ , $3 \times (-2) = -6$ .

(b) $ab \times a = a^2b$ , $ab \times (-b) = -ab^2$ , $1 \times a = a$ , $1 \times (-b) = -b$ .

(a) $x^3 - 2x^2 + 3x - 6$ ; (b) $a^2b - ab^2 + a - b$

Question 9

Show that $(n + 2)(n + 5) - (n + 1)(n + 6) = 4$ for every value of $n$ , and check it with $n = 3$ .

ShowHide solution

$(n + 2)(n + 5) = n^2 + 7n + 10$ and $(n + 1)(n + 6) = n^2 + 7n + 6\text{.}$

$(n^2 + 7n + 10) - (n^2 + 7n + 6) = 4$

With $n = 3$ : $5 \times 8 - 4 \times 9 = 40 - 36 = 4\text{.}$ ✓

The difference is $4$ for every $n$ .

Question 10

A photograph $x$ cm by $(x + 4)$ cm is mounted on a card so that there is a border $3$ cm wide all round it. (a) Find, in its simplest form, the area of the border. (b) Find this area when $x = 10$ .

ShowHide solution

(a) The card is $(x + 6)$ cm by $(x + 10)$ cm.

Border $= (x + 6)(x + 10) - x(x + 4) = x^2 + 16x + 60 - x^2 - 4x = 12x + 60$

(b) $12(10) + 60 = 180$ . Check: the card is $16$ cm by $20$ cm, area $320$ cm²; the photograph is $10$ cm by $14$ cm, area $140$ cm²; $320 - 140 = 180$ . ✓

(a) $(12x + 60)$ cm²; (b) $180$ cm²

Question 11

A box is a cuboid $x$ cm wide, $(x + 3)$ cm long and $(x - 1)$ cm high. Find its volume in expanded form, and check your answer when $x = 4$ .

ShowHide solution

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

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

When $x = 4$ : $4 \times 7 \times 3 = 84$ , and $64 + 32 - 12 = 84$ . ✓

$(x^3 + 2x^2 - 3x)$ cm³

Question 12

Jun Hao wrote $(x + 1)(x + 2) - (x + 3)(x - 1) = x^2 + 3x + 2 - x^2 + 2x - 3 = 5x - 1\text{.}$ Check his answer with $x = 2$ , find his mistake and correct it.

ShowHide solution

With $x = 2$ : $3 \times 4 - 5 \times 1 = 7$ , but $5(2) - 1 = 9$ . So the answer is wrong.

$(x + 3)(x - 1) = x^2 + 2x - 3\text{.}$ Jun Hao changed only the sign of $x^2$ . The minus must change every sign:

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

Check: $2 + 5 = 7$ . ✓

$x + 5$

Key Points

Every term of one bracket multiplies every term of the other; two terms by three terms give six products.
Write the letters of each term in alphabetical order, and collect like terms such as $xy$ and $yx$ .
With a number or term in front of two brackets, expand the brackets first, then multiply every term by it once.
Keep an expansion that follows a minus sign in a bracket; the minus changes every sign inside.
To find one coefficient, add only the products that give that term.
Check by substituting a number into the question and the answer.

Expanding Longer Products and Simplifying

Before you start

Brackets with Different Letters

Example 1: Four different letters

Example 2: In a frame

Two Letters in Each Bracket

Example 3: Terms in p2p^2p2, pqpqpq and q2q^2q2

A Bracket with Three Terms

More terms, more products

Example 4: A frame with three columns

Example 5: Three terms in the first bracket

Example 6: One coefficient only

A Number or a Term in Front of Two Brackets

Example 7: A number in front

Example 8: A letter in front

Sums and Differences of Products

A minus sign before a product

Example 9: The difference of two products

Example 10: A term times a bracket, minus two brackets

Example 11: Always the same

Areas

Example 12: A path round a lawn

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: Terms in $p^2$ , $pq$ and $q^2$