International Baccalaureate · MYP 3

Expansion and Factorisation

⌚ ~154 min ▢ 17 blocks ✓ 74 exercises ➣ Prerequisites : Algebraic Expressions and Fractions

I Expanding

I.1 The Distributive Law

The large rectangle below is split into two smaller rectangles. Find the total area of the large rectangle in two different ways.

Answer

Method 1: Sum of the parts
The total area is the sum of the areas of the two smaller rectangles. $$\text{Total Area} = \textcolor{colordef}{\text{Area 1}} + \textcolor{colorprop}{\text{Area 2}} = \textcolor{colordef}{ab} + \textcolor{colorprop}{ac}$$
Method 2: Area of the whole
The total length of the base is $b+c$ and the height is $a$. $$\text{Total Area} = \textcolor{olive}{a(b+c)}$$

Since both methods calculate the same total area, the two expressions must be equal. This gives us the identity: $$\textcolor{olive}{a(b+c)} = \textcolor{colordef}{ab} + \textcolor{colorprop}{ac}$$ This important rule is known as the distributive law.

Proposition — Distributive Law

For any numbers $a$, $b$ and $c$, multiplication is distributive over addition and subtraction:

Addition:
Subtraction:

The rectangles show positive lengths only; the law holds for every number, negatives included.

Example

Show that $2(l + w) = 2l + 2w$.

Answer

Method — Sum or Product?

The distributive identities convert between two shapes of expression. To decide which shape an expression has, look at its last operation — the one performed last when computing:

in $3(x+2)$, one computes the bracket, then multiplies: the last operation is a multiplication — the expression is a product;
in $3x+2$, one computes $3x$, then adds $2$: the last operation is an addition — the expression is a sum.

Neither brackets nor invisible signs decide the shape: only the last operation does.

Example

Sum or product? Decide for each expression: $$5(x+1), \qquad 5x+1, \qquad 2(x+3)+1$$

Answer

$5(x+1)$: compute the bracket, then multiply by $5$ — a product.
$5x+1$: compute $5x$, then add $1$ — a sum.
$2(x+3)+1$: compute the bracket, multiply by $2$, then add $1$ — the brackets suggest a product, but the last operation is the addition of $1$: a sum.

Definition — Expanding

Expanding is the process of using the distributive law to write a product with brackets as a sum (or difference) of terms.

Example

Expand $2(2x + 3)$.

Answer

Example

Expand and simplify: $x(2x+3)$.

Answer

$$\begin{aligned}[t] x(2x + 3) &= x \times 2x + x \times 3 && {\scriptstyle\text{(distributivity)}}\\ &= \textcolor{colordef}{2x^2 + 3x} && {\scriptstyle(x \times 2x = 2x^2)} \end{aligned}$$

Skills to practice

Recognising a Sum or a Product
Expanding with Addition: Level 1
Expanding with Addition: Level 2
Expanding with Addition: Level 3
Expanding with Subtraction: Level 1
Expanding with Subtraction: Level 2
Expanding with Subtraction: Level 3

I.2 The Opposite of an Expression

Method — Taking the Opposite of an Expression

The opposite of an expression is the expression multiplied by $(-1)$. To write it without brackets:

Write the expression in brackets, preceded by a $-$ sign.
Change the sign of every term: this is exactly distributing the factor $(-1)$ over the brackets.
Check with a value: the expression and its opposite must give opposite results.

Subtracting an expression amounts to adding its opposite: $A - (B) = A + (-B)$. This is why, in a subtraction, every term of the brackets changes sign.

Example

Write the opposite of $5 - 4x$ without brackets.

Answer

$$\begin{aligned}[t]-(5 - 4x) &= (-1) \times 5 + (-1) \times (-4x) && \text{(Distribute the factor $-1$)} \\ &= -5 + 4x && \text{(Each term changes sign)}\end{aligned}$$ Check with $x = 1$: the expression gives $5 - 4 = 1$, and $-5 + 4 = -1$ — opposite results.

Example

Expand and simplify $-3(x+2)$, then $6-2(x-4)$.

Answer

Taking the opposite is the case $-1$; any other negative factor is distributed in the same way — the factor carries its sign onto every term of the bracket. $$\begin{aligned}[t]-3(x+2) &= (-3) \times x + (-3) \times 2 && \text{(Distribute the factor $-3$)} \\ &= \textcolor{colorprop}{-3x-6} && \text{(Each term is multiplied by $-3$)}\end{aligned}$$ $$\begin{aligned}[t]6-2(x-4) &= 6 + (-2) \times x + (-2) \times (-4) && \text{(Distribute the factor $-2$)} \\ &= 6 - 2x + 8 && \text{(A negative times a negative is positive)} \\ &= \textcolor{colorprop}{14 - 2x} && \text{(Combine like terms)}\end{aligned}$$ Checking with $x = 1$: $-3 \times 3 = -9$ and $-3-6 = -9$; $6 - 2 \times (-3) = 12$ and $14 - 2 = 12$.

Skills to practice

Taking the Opposite: Level 1
Taking the Opposite: Level 2

II Factorising

II.1 Factorising with a Common Factor

The common factor laws are the reverse of the distributive laws: instead of expanding brackets, we look for a common factor to “take out” and introduce brackets.

Report an error

Proposition — Common Factor Laws

For any numbers $a$, $b$ and $c$: $\textcolor{colorprop}{a}b+\textcolor{colorprop}{a}c=\textcolor{colorprop}{a}(b+c)\quad$ and $\quad\textcolor{colorprop}{a}b-\textcolor{colorprop}{a}c=\textcolor{colorprop}{a}(b-c)$

Example

Factorise $2x+2$.

Answer

$$\begin{aligned} 2x+2&=\textcolor{colorprop}{2}\times x +\textcolor{colorprop}{2}\times 1\\ &=\textcolor{colorprop}{2}(x+1). \end{aligned}$$

Example

Factorise: $x^2 + 6x$.

Answer

$$\begin{aligned}[t] x^2 + 6x &= \textcolor{colorprop}{x} \times x + \textcolor{colorprop}{x} \times 6 && {\scriptstyle\text{(the common factor is } \textcolor{colorprop}{x})}\\ &= \textcolor{colorprop}{x}(x + 6) \end{aligned}$$

Method — Factorising with a Common Bracket

A common factor is not always a number or a letter: it can be a whole bracket appearing in every term.

Spot the bracket that appears in each term.
Write this bracket in front, and collect in a second bracket what multiplies it in each term.
Simplify the second bracket.

When the second product is subtracted, the minus sign applies to everything that enters the second bracket: in $(x+1)(2x+3) - (x+1)(x-4)$, the second bracket is $(2x+3) - (x-4)$.
A term made of the common bracket alone is multiplied by $1$: in $(x+3)(2x-1) + (x+3)$, the second term is $(x+3) \times 1$, so the second bracket is $(2x-1) + 1$.

Example

Factorise $(x+2)(3x-1) + (x+2)(x+5)$.

Answer

The bracket $\textcolor{colorprop}{(x+2)}$ appears in both terms: it is the common factor. $$\begin{aligned}[t] &(x+2)(3x-1) + (x+2)(x+5) \\ &= \textcolor{colorprop}{(x+2)}\left[(3x-1) + (x+5)\right] && {\scriptstyle\text{(common bracket)}} \\ &= \textcolor{colorprop}{(x+2)}(3x-1+x+5) \\ &= \textcolor{colorprop}{(x+2)}(4x+4). \end{aligned}$$ The second bracket still holds the common factor $4$, so the complete factorisation is $4(x+2)(x+1)$.

Skills to practice

Factorising with a Common Factor: Level 1
Factorising with a Common Factor: Level 2
Factorising with a Common Bracket

III Proving with Algebra

III.1 Proving by Expanding and Factorising

Discover

Zoe claims: “The sum of two odd numbers is always even.”
Check her claim with $3 + 5$, with $7 + 13$, and with $9 + 21$. These three checks agree with Zoe — but can you be sure the claim is true for all odd numbers, without testing them all?

Definition — Even and Odd Numbers in Algebraic Form

For any whole number $n$:

the number $2n$ is even — it is twice $n$; every even number can be written in this form;
the number $2n + 1$ is odd — it is an even number plus one; every odd number can be written in this form.

Method — Proving by Expanding and Factorising

Testing examples can only make a claim plausible; to prove it for all numbers:

Write the numbers in algebraic form: $2n$ for an even number, $2n+1$ for an odd number, $n$ and $n+1$ for consecutive numbers… Use different letters, such as $n$ and $m$, for numbers chosen independently.
Compute with the letters (expand, simplify, factorise).
Read the form of the result — $2(\ldots)$ is twice a whole number, $3(\ldots)$ a multiple of $3$… — and connect it to the claim: this form proves the property for every whole number.

Example

Prove Zoe's claim: the sum of two odd numbers is always even.

Answer

Let $2n + 1$ and $2m + 1$ be two odd numbers ($n$ and $m$ whole numbers — two different letters, because the two odd numbers are chosen independently). $$\begin{aligned}[t] (2n + 1) + (2m + 1) &= 2n + 2m + 2 && {\scriptstyle\text{(remove the brackets and reorder)}}\\ &= 2(n + m + 1) && {\scriptstyle\text{(common factor } 2)} \end{aligned}$$ The sum is twice the whole number $n + m + 1$: it is even, for every choice of the two odd numbers. Zoe is correct.

Disproving with a Counter-Example

To prove that a general claim is false, one single counter-example is enough. For instance, “every odd number is prime” is false, because $9$ is odd and $9 = 3 \times 3$ is not prime. No amount of successful tests can prove a claim about infinitely many numbers — but one counter-example disproves it.

Report an error

Skills to practice

Proving Properties of Even and Odd Numbers
Proving with Factorisation

IV Going further

Skills to practice

Reasoning and Problem-Solving
Solving Problems in Context
Solving Problems with Algebra
Quiz