Expand and Simplify

Collecting like terms, multiplying terms, expanding brackets and simplifying, three brackets and simple algebraic proof.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 160 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What Expanding and Simplifying Mean

Simplify means write it shorter. Expand means get rid of the brackets.

Example

Think of 4 apples + 2 bananas + 3 apples. You can put the apples together: 7 apples + 2 bananas. The apples and the bananas stay apart.

Rule

Like terms have exactly the same letters and powers. $3 x$ and $5 x$ are like terms. $3 x$ and $3 x^{2}$ are not.

Key idea

Almost every question in this lesson ends the same way: collect the like terms.

Key idea

The word algebra comes from al-jabr, in the title of a book written in Baghdad around 820 by al-Khwarizmi. Al-jabr means restoring, and al-muqabala, his other method, means balancing: taking the same thing from both sides. The word algorithm comes from his name.

Collecting Like Terms

This is the most common question: Simplify a line of added and subtracted terms.

Rule

Each term owns the sign in front of it. Move the term, and its sign goes with it.

Worked example

1
$4 a + 2 b - a + 3 b$ — The question. There are two kinds of term, a's and b's, so the answer will have two terms.
2
$4 a + 2 b - a + 3 b$ — Mark each a term and each b term. Include the sign in front, because it belongs to the term.
3
$= 4 a - a + 2 b + 3 b$ — Move like terms side by side. The −a keeps its minus when it moves.
4
$= 3 a + 5 b$ — 4a − a = 3a because −a means take away one a. 2b + 3b = 5b.

Common mistake

Reading $- a$ as 0. On its own a letter means one of it: $- a$ is $- 1 a$ .

With powers. An $x^{2}$ term and an $x$ term are different kinds, just like a and b. Collect each kind on its own.

Worked example

1
$5 x^{2} + 3 x - 2 x^{2} - 7 x$ — The question. Two kinds of term: x² terms and x terms.
2
$= 5 x^{2} - 2 x^{2} + 3 x - 7 x$ — Put like terms side by side, each with its own sign.
3
$= 3 x^{2} - 4 x$ — 5 − 2 = 3, so 3x². 3 − 7 = −4, so −4x. The power stays the same: adding never changes it.

Common mistake

Joining $3 x^{2} - 4 x$ into one term. They are different kinds, so $3 x^{2} - 4 x$ is already the final answer.

Multiplying Terms

When the terms are multiplied, the rule changes. Order does not matter when you multiply, so group the numbers and group the letters.

Worked example

1
$3 p \times 4 q$ — The question. A times sign between two terms means multiply everything in them.
2
$= 3 \times 4 \times p \times q$ — Reorder: numbers together, letters together. You may, because order does not matter in multiplying.
3
$= 12 pq$ — 3 × 4 = 12, and p × q is written pq. Write the number first, then the letters in alphabet order.

Adding

3 c + 5 c = 8 c

Count them: 8 lots of c.

Multiplying

3 c \times 5 c = 15 c^{2}

c times c is c².

Common mistake

Writing $3 c \times 5 c = 15 c$ . The two c's multiply too, so you get $c^{2}$ .

Expanding Brackets, Then Simplifying

The number or term in front multiplies every term inside the bracket. Then collect.

Worked example

1
$2 x (3 x + 4) - 3 (x - 2)$ — The question. Two brackets, each with a term in front. Expand one at a time, then collect.
2
$= 6 x^{2} + 8 x - 3 (x - 2)$ — 2x times each term: 2x × 3x = 6x², because x × x = x², and 2x × 4 = 8x.
3
$= 6 x^{2} + 8 x - 3 x + 6$ — −3 times each term: −3 × x = −3x, and −3 × (−2) = +6, because minus times minus is plus.
4
$= 6 x^{2} + 5 x + 6$ — Collect the x terms: 8x − 3x = 5x. The 6x² and the 6 have no partners, so they stay.

Common mistake

Writing $- 3 (x - 2) = - 3 x - 6$ . The minus multiplies the −2 as well, and minus times minus is plus.

A power in front. Exams often put a power outside the bracket. Multiply the numbers, and add the powers of the same letter.

Worked example

1
$4 p^{2} (3 p^{3} - 2)$ — The question. 4p² multiplies both terms inside.
2
$4 p^{2} \times 3 p^{3} = 12 p^{5}$ — 4 × 3 = 12, and $p^{2} \times p^{3} = p^{5}$ , because 2 p's times 3 p's makes 5 p's.
3
$4 p^{2} \times (- 2) = - 8 p^{2}$ — 4 × (−2) = −8. There is no p in −2, so the p² stays as it is.
4
$12 p^{5} - 8 p^{2}$ — Write both parts. They have different powers, so they are not like terms and cannot be collected.

Common mistake

Writing $p^{2} \times p^{3} = p^{6}$ . When you multiply, you add the powers: 2 + 3 = 5.

Three Brackets (Extended/Higher)

Papers ask you to expand three brackets for 2 or 3 marks. Never do all three at once: multiply two at a time.

Worked example

1
$(x + 2) (x - 3) (2 x + 1)$ — The question. Start with the first two brackets and leave (2x + 1) for later.
2
$(x + 2) (x - 3) = x^{2} - 3 x + 2 x - 6$ — Each term in the first bracket times each in the second: x × x, x × (−3), 2 × x, 2 × (−3).
3
$= x^{2} - x - 6$ — Collect: −3x + 2x = −x. Doing this now means fewer terms in the next step.
4
$(x^{2} - x - 6) (2 x + 1)$ — Now multiply that answer by the last bracket.
5
$x^{2} \times (2 x + 1) = 2 x^{3} + x^{2}$ — The first term times both: x² × 2x = 2x³ and x² × 1 = x².
6
$- x \times (2 x + 1) = - 2 x^{2} - x$ — The second term, with its minus: −x × 2x = −2x² and −x × 1 = −x.
7
$- 6 \times (2 x + 1) = - 12 x - 6$ — The third term: −6 × 2x = −12x and −6 × 1 = −6.
8
$= 2 x^{3} + x^{2} - 2 x^{2} - x - 12 x - 6$ — Write all six terms in one line. 3 terms times 2 terms makes 6.
9
$= 2 x^{3} - x^{2} - 13 x - 6$ — Collect: x² − 2x² = −x², and −x − 12x = −13x.

Key idea

Count before you collect: 3 terms times 2 terms makes 6 terms. If you have fewer, one is missing.

Find a, b and c (Extended/Higher)

Some questions give an identity, written with $\equiv$ , and ask you to find the letters. $\equiv$ means both sides are equal for every x. So the number in front of each power must match on both sides.

Rule

Expand the side with the brackets. Then match $x^{2}$ with $x^{2}$ , x with x, and number with number.

A squared bracket. Textbooks print the general result for squaring a bracket:

Formula

(a x + b)^{2} \equiv a^{2} x^{2} + 2 ab x + b^{2}

$a$ is the number in front of x in the bracket and $b$ is the number on its own, each with its sign. Square a, double a × b, square b.

Worked example

1
$(a x + b)^{2} \equiv 9 x^{2} - 24 x + c, a > 0$ — The question. Match each part of the formula with the right side.
2
$a^{2} = 9$ — Match the x² terms: a² is the number in front of x².
3
$a = 3$ — a could be 3 or −3, since both square to 9. The question says a > 0, so −3 is impossible. Reject it.
4
$2 ab = - 24$ — Match the x terms: 2ab is the number in front of x.
5
$2 \times 3 \times b = - 24, so b = - 4$ — Put a = 3 in: 6b = −24. Divide both sides by 6.
6
$c = b^{2} = (- 4)^{2} = 16$ — Match the numbers on their own: c is b². Brackets on −4: minus times minus is plus.

Three brackets. Find the letter inside the brackets first, using the numbers on their own. Then expand.

Worked example

1
$(x + a) (x + 3) (2 x - 1) \equiv 2 x^{3} + b x^{2} + c x + 12$ — The question. Find a, b and c.
2
$a \times 3 \times (- 1) = 12$ — The number on its own comes only from multiplying the three numbers in the brackets.
3
$- 3 a = 12, so a = - 4$ — 3 × (−1) = −3. Divide both sides by −3: 12 ÷ (−3) = −4.
4
$(x - 4) (x + 3) = x^{2} - x - 12$ — Put a = −4 in and expand two brackets: 3x − 4x = −x, and (−4) × 3 = −12.
5
$(x^{2} - x - 12) (2 x - 1) = 2 x^{3} - x^{2} - 2 x^{2} + x - 24 x + 12$ — Each of the 3 terms times each of the 2: six terms.
6
$= 2 x^{3} - 3 x^{2} - 23 x + 12$ — Collect: −x² − 2x² = −3x², and x − 24x = −23x.
7
$b = - 3, c = - 23$ — Match: b is the number in front of x², c the number in front of x. The 2x³ matches too, which checks the expanding.

Common mistake

Keeping both answers for a when the question says a > 0. A condition like that is there to rule one of them out.

Prove It with Algebra (Extended/Higher)

“Prove that …” questions want algebra, not examples. Write the numbers with a letter, expand, then show the answer has the property.

You need	Write it as
any whole number	$n$
an even number	$2 n$
an odd number	$2 n + 1$
numbers in a row	$n, n + 1, n + 2$
even numbers in a row	$2 n, 2 n + 2, 2 n + 4$
odd numbers in a row	$2 n + 1, 2 n + 3$
two different odd numbers	$2 m + 1, 2 n + 1$

Prove that the difference between the squares of two odd numbers in a row is a multiple of 8.

Worked example

1
$2 n + 1, 2 n + 3$ — Two odd numbers in a row. Odd numbers are 2 apart, so add 2 to get the next one.
2
$(2 n + 3)^{2} - (2 n + 1)^{2}$ — The difference of their squares: the bigger square take away the smaller.
3
$= 4 n^{2} + 12 n + 9 - (4 n^{2} + 4 n + 1)$ — Expand each square. Keep the bracket, because the minus belongs to all of it.
4
$= 4 n^{2} + 12 n + 9 - 4 n^{2} - 4 n - 1$ — Open the bracket: the minus changes the sign of every term inside.
5
$= 8 n + 8$ — Collect: 4n² − 4n² = 0, 12n − 4n = 8n, and 9 − 1 = 8.
6
$= 8 (n + 1)$ — Take out 8. This is 8 times a whole number, so it is always a multiple of 8.

To prove it is odd, get the answer into the form $2 (\dots) + 1$ : an even number plus 1. Prove that the square of any odd number is odd.

Worked example

1
$(2 n + 1)^{2}$ — The square of any odd number.
2
$= 4 n^{2} + 4 n + 1$ — Expand: 2n × 2n = 4n², 2n × 1 twice gives 4n, and 1 × 1 = 1.
3
$= 2 (2 n^{2} + 2 n) + 1$ — Take 2 out of the first two terms only. 2 times a whole number is even.
4
$even + 1 = odd$ — An even number plus 1 is always odd, so the square is odd.

Common mistake

Trying a few numbers (3 and 5, 7 and 9) and stopping. Examples show it; only the algebra proves it for every number.

Common mistake

Using 2n + 1 for two different odd numbers. That proves it only when both numbers are the same. Use 2m + 1 and 2n + 1.

Key idea

End with a sentence that says why, like “8(n + 1) is 8 times a whole number, so it is a multiple of 8”. The last mark is for that sentence.

Which Method? Read the Question

The command word tells you what to do.

You see	You do
“Simplify”, terms added or taken away	Collect like terms, signs included
“Simplify”, terms multiplied	Numbers together, letters together
“Expand”	Multiply every term inside the bracket
A power outside the bracket	Multiply the numbers, add the powers
“Expand and simplify”	Expand, then collect like terms
Three brackets	Two at a time, collect, then the third
≡ and “find a, b and c”	Expand, then match each power
“Prove” or “Show that”	Use n, expand, then factorise

Key idea

Single-bracket questions are 1–2 marks; three brackets and proofs are 3. Write the expanded line before you collect: if the final answer slips, that line can still earn a mark. The last line must be fully collected to score the final mark.

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