Expanding Two Brackets

Expanding and simplifying two brackets: four products, signs, back-to-front brackets, two-letter terms and squared brackets.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 51 real past-paper questions below

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Two Brackets, Four Products

Every question here says expand and simplify for 2 marks. Each term in the first bracket multiplies each term in the second.

Rule

2 terms times 2 terms makes 4 products. Then collect the two middle terms.

Key idea

$(x + 2)$ invited $(x + 3)$ to dinner. Four guests turned up: $x^{2}$ , $3 x$ , $2 x$ and $6$ .

Expanding Two Brackets

Every term in the first bracket multiplies every term in the second. Go in order, one product at a time.

Worked example

1
$(x + 6) (x - 4)$ — The question. Two terms times two terms, so expect 4 products.
2
$x \times x = x^{2}$ — First × first. x times x is x squared.
3
$x \times (- 4) = - 4 x$ — First × second. The −4 keeps its minus, so this product is negative.
4
$6 \times x = 6 x$ — Now the +6 takes its turn: +6 times the first term, x.
5
$6 \times (- 4) = - 24$ — Last product: +6 times −4. Plus times minus is minus.
6
$= x^{2} - 4 x + 6 x - 24$ — Write all four products in one line, each with its own sign.
7
$= x^{2} + 2 x - 24$ — Collect the two x terms: −4x + 6x = 2x. The x² and −24 have no partner, so they stay.

Key idea

Always show the four-product line. It earns a mark for working if at least 3 of the 4 terms are right, even if your final answer slips.

Common mistake

Stopping at $x^{2} - 24$ . That is only 2 of the 4 products. The middle terms are missing.

Watch the Signs

Keep each sign glued to its number. Minus × plus is minus; minus × minus is plus.

Worked example

1
$(m - 7) (m + 3)$ — The question. The −7 takes its minus into both of its products.
2
$m \times m = m^{2}$ — First × first.
3
$m \times 3 = 3 m$ — First × second. Both are positive, so the product is positive.
4
$(- 7) \times m = - 7 m$ — Now −7 times m. Minus times plus is minus.
5
$(- 7) \times 3 = - 21$ — −7 times +3. Minus times plus is minus again.
6
$= m^{2} + 3 m - 7 m - 21$ — All four products in one line.
7
$= m^{2} - 4 m - 21$ — Collect: 3m − 7m = −4m.

Both minus: now the last product turns positive.

Worked example

1
$(p - 5) (p - 6)$ — The question. Both brackets have a minus.
2
$p \times p = p^{2}$ — First × first.
3
$p \times (- 6) = - 6 p$ — First × second. Plus times minus is minus.
4
$(- 5) \times p = - 5 p$ — Minus times plus is minus.
5
$(- 5) \times (- 6) = + 30$ — Minus times minus is plus, so this is +30, not −30.
6
$= p^{2} - 6 p - 5 p + 30$ — All four products in one line.
7
$= p^{2} - 11 p + 30$ — Collect: −6p − 5p = −11p. Two minuses add up to a bigger minus.

Common mistake

Writing −30 at the end. $(- 5) \times (- 6) = + 30$ . Put a negative number in brackets so you see both minus signs.

When a Bracket Is Back to Front

Some papers write the letter second, like $(5 - y)$ . Nothing changes: still four products. Watch the y × −y.

Worked example

1
$(y + 3) (5 - y)$ — The question. The second bracket starts with the number, and its y is negative.
2
$y \times 5 = 5 y$ — First × first. Write the number before the letter.
3
$y \times (- y) = - y^{2}$ — y × y is y², and the minus stays: −y². This is the term students lose.
4
$3 \times 5 = 15$ — Now the +3 times 5.
5
$3 \times (- y) = - 3 y$ — The +3 times −y. Plus times minus is minus.
6
$= 5 y - y^{2} + 15 - 3 y$ — All four products in one line.
7
$= 15 + 2 y - y^{2}$ — Collect: 5y − 3y = 2y. The −y² and 15 have no partner.

Key idea

Any order is fine for the mark: $- y^{2} + 2 y + 15$ is the same answer.

A Number in Front of x

When x has a number in front, multiply the numbers and then the letters. So $4 x \times 2 x$ is 4 × 2 = 8 and x × x = x², giving 8x².

Worked example

1
$(4 x + 3) (2 x - 5)$ — The question. Both x terms have a number in front.
2
$4 x \times 2 x = 8 x^{2}$ — Multiply the numbers, 4 × 2 = 8, then the letters, x × x = x².
3
$4 x \times (- 5) = - 20 x$ — 4 × −5 = −20, and the x stays.
4
$3 \times 2 x = 6 x$ — Now the +3 times 2x: 3 × 2 = 6.
5
$3 \times (- 5) = - 15$ — The +3 times −5. Plus times minus is minus.
6
$= 8 x^{2} - 20 x + 6 x - 15$ — All four products in one line.
7
$= 8 x^{2} - 14 x - 15$ — Collect: −20x + 6x = −14x.

Common mistake

Adding the numbers: $4 x \times 2 x = 6 x^{2}$ . Products multiply: $8 x^{2}$ .

Two Letters in the Brackets

Some brackets hold two letters, like x and y. Multiply the numbers, then write the letters side by side.

Worked example

1
$(3 x - y) (x + 3 y)$ — The question. Still four products.
2
$3 x \times x = 3 x^{2}$ — 3 × 1 = 3, and x × x = x².
3
$3 x \times 3 y = 9 x y$ — 3 × 3 = 9, and x times y is written xy.
4
$(- y) \times x = - x y$ — Minus times plus is minus. y × x is the same as x × y, so write it xy.
5
$(- y) \times 3 y = - 3 y^{2}$ — −1 × 3 = −3, and y × y = y².
6
$= 3 x^{2} + 9 x y - x y - 3 y^{2}$ — All four products in one line.
7
$= 3 x^{2} + 8 x y - 3 y^{2}$ — 9xy − xy = 8xy. The x² and y² are different terms, so they stay apart.

A letter in place of a number. Treat it just like a number: $2 a \times 3 x = 6 a x$ .

Worked example

1
$(x + 2 a) (3 x - a)$ — The question. The letter a stands for a number, so +2a is just a term.
2
$x \times 3 x = 3 x^{2}$ — First × first.
3
$x \times (- a) = - a x$ — Plus times minus is minus. Write the letters in alphabetical order: ax.
4
$2 a \times 3 x = 6 a x$ — 2 × 3 = 6, and the letters a and x sit side by side.
5
$2 a \times (- a) = - 2 a^{2}$ — 2 × −1 = −2, and a × a = a².
6
$= 3 x^{2} - a x + 6 a x - 2 a^{2}$ — All four products in one line.
7
$= 3 x^{2} + 5 a x - 2 a^{2}$ — Collect: −ax + 6ax = 5ax.

Common mistake

Treating xy and yx as different terms. They are the same, so 9xy − yx = 8xy.

Squaring a Bracket

A squared bracket means the bracket times itself. Textbooks print the result as two formulas:

Formula

(a + b)^{2} = a^{2} + 2 ab + b^{2}

Formula

(a - b)^{2} = a^{2} - 2 ab + b^{2}

$a$ is the first term in the bracket and $b$ is the second. Square the first, double the two multiplied, square the second. The sign in the middle copies the bracket; the end is always +.

Worked example

1
$(x + 7)^{2} = (x + 7) (x + 7)$ — Squared means times itself, so write the bracket twice.
2
$= x^{2} + 7 x + 7 x + 49$ — The four products: x × x, x × 7, 7 × x and 7 × 7. The two middle ones are equal.
3
$= x^{2} + 14 x + 49$ — 7x + 7x = 14x. That is the $2 ab$ in the formula: 2 × x × 7.

Worked example

1
$(3 x - 2)^{2}$ — The question. Here a = 3x and b = 2, with a minus, so use the second formula.
2
$a^{2} = (3 x)^{2} = 9 x^{2}$ — Square the whole first term: 3² = 9 and x² stays.
3
$2 ab = 2 \times 3 x \times 2 = 12 x$ — Double the two terms multiplied. It goes in with a minus, like the bracket.
4
$b^{2} = 2^{2} = 4$ — Square the second term. The end is always +.
5
$= 9 x^{2} - 12 x + 4$ — Put the three parts together: first squared, minus the middle, plus the end.

Common mistake

Writing $(x + 7)^{2} = x^{2} + 49$ . The middle term 14x is missing. Also, $(3 x)^{2}$ is 9x², not 3x².

Find the Letters (Extended/Higher)

Some questions give an answer with letters in it and ask you to find them. Expand the side with brackets, then match the two sides term by term.

Worked example

1
$(x + a)^{2} = x^{2} + 14 x + b$ — The question. Find a and b.
2
$x^{2} + 2 a x + a^{2} = x^{2} + 14 x + b$ — Expand the left with the square formula: the middle is 2ax and the end is a².
3
$2 a = 14$ — The x terms must match, so 2a = 14.
4
$a = 7$ — Divide both sides by 2.
5
$b = a^{2} = 7^{2} = 49$ — The numbers on their own must match too, so b = a².
6
$(x + 7)^{2} = x^{2} + 14 x + 49$ — Check by expanding. It matches, so a = 7 and b = 49.

Worked example

1
$(x - 3)^{2} + k = x^{2} - p x + 2$ — The question. Find p and k.
2
$x^{2} - 6 x + 9 + k = x^{2} - p x + 2$ — Expand the square: −3x twice makes −6x, and (−3)² = +9.
3
$- 6 = - p$ — Match the x terms: −6x = −px.
4
$p = 6$ — Both sides have a minus, so p = 6.
5
$9 + k = 2$ — Match the numbers on their own: 9 + k = 2.
6
$k = 2 - 9 = - 7$ — Take 9 from both sides.

Key idea

Each letter usually earns its own mark. If you find one and not the other, still write the one you found.

Which Method? Read the Question

Count the terms, then check the signs.

You see	You do
Two brackets	4 products, collect the middle
A bracket squared	write it twice, or use a² + 2ab + b²
Letters to find on both sides	expand, then match term by term
A rectangle with sides in brackets	area = one side × the other, then expand
A number × one bracket	see Expand and Simplify
Three brackets	two at a time (Expand and Simplify)
Make it a product	see Factorising Quadratic Expressions

Brackets hidden in a shape. A rectangle is $(x + 4)$ cm long and $(x - 1)$ cm wide. Write its area in the form $x^{2} + a x + b$ .

Worked example

1
$area = (x + 4) (x - 1)$ — Area of a rectangle = length × width. Writing this product can already earn a mark.
2
$= x^{2} - x + 4 x - 4$ — The four products: x × x, x × (−1), 4 × x and 4 × (−1).
3
$= x^{2} + 3 x - 4$ — Collect: −x + 4x = 3x. So a = 3 and b = −4.

Key idea

Check: put x = 1 into the question and your answer. Both should give the same number.

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