2.2 - Expanding brackets and common factors

Expanding turns a product into a sum; factorising reverses that change. You will multiply through brackets, find common factors, and expand two simple linear factors. The final part extends the method to more general products at Higher.

Distributing a single term

Can you calculate $99\times101$ without long multiplication? Brackets will let us turn it into a calculation built around 100. First we need a rule that keeps every term in a product.

You already know

Multiply numerical coefficients and combine matching letter factors. Collect like terms only when their variable parts match, keeping each term's sign.

Three bags containing $x+2$ counters each contain $3x+6$ counters in total. Both the $x$ counters and the 2 extra counters occur three times. This is the distributive law.

Distributive law

a(b+c)=ab+ac,\qquad a(b-c)=ab-ac

To expand, carry out the multiplication so the brackets disappear. These identities hold for all real values.

Not in either tier's formulae sheet: learn them.

The multiplier may contain a letter or be negative. Multiply it by every term, not just the first one.

Worked example 1: a negative multiplier

Question. Expand $-3x(2x-5)$ .

Plan. Multiply $-3x$ by each signed term inside the bracket.

\begin{aligned} -3x(2x-5)&=(-3x)(2x)+(-3x)(-5) &&\text{distribute}\\ &=-6x^2+15x &&\text{multiply each term} \end{aligned}

Why this step? The negative multiplier applies to the $-5$ as well as the $2x$ , so the second product is positive.

Answer. $-6x^2+15x$ .

Check. At $x=2$ , the product is $-6(-1)=6$ , and the answer is $-24+30=6$ .

Change only the sign inside the bracket. The first product stays the same; the second changes sign.

Worked example 2: changing an inside sign

Question. Expand $-3x(2x+5)$ .

Plan. Distribute the same negative multiplier over both terms.

\begin{aligned} -3x(2x+5)&=(-3x)(2x)+(-3x)(5) &&\text{distribute}\\ &=-6x^2-15x &&\text{multiply with signs} \end{aligned}

Why this step? A negative times a positive is negative, so both terms are now negative.

Answer. $-6x^2-15x$ .

Check. At $x=2$ , $-6(9)=-54$ , matching $-24-30=-54$ .

For separate brackets in a sum, expand each using its own multiplier before collecting. For example,

5(2a-3)-2(a+4)=10a-15-2a-8=8a-23

A minus before a bracket means a multiplier of $-1$ . It changes every sign. Addition between brackets does not mean that the brackets should be multiplied together.

Watch out

The line $-2(a+4)=-2a+8$ is wrong: $-2$ multiplies the positive 4 too. The correct line is $-2(a+4)=-2a-8$ .

Try the same bracket with a different multiplier. Keep the two products separate until you have checked their signs.

Taking out common factors

Factorising runs the distributive law backwards. Instead of writing two products separately, show the multiplier they share.

Common factor

A common factor divides every term. To factorise an expression is to write it as a product. To factorise fully, take out the highest common numerical factor and every shared letter factor, leaving no common factor inside the bracket.

Choose the lowest power of each letter that occurs in every term. A letter missing from any term cannot be taken out of the whole expression. Divide each term by the chosen factor to find what remains inside. This is a use of the distributive law, which is not supplied in either tier's formulae sheet.

Worked example 3: choosing the full common factor

Question. Factorise fully $18a^2b-12ab^2$ .

Plan. Find the numerical HCF and lowest shared letter powers, then divide both terms by that factor.

\begin{aligned} \text{HCF}&=6ab &&\text{shared in both terms}\\ 18a^2b-12ab^2&=6ab(3a-2b) &&\text{divide each term by }6ab \end{aligned}

Why this step? Both terms contain $a$ and $b$ , but neither contains two of both letters.

Answer. $6ab(3a-2b)$ .

Check. Expanding gives $18a^2b-12ab^2$ , the original expression term by term.

Now change only the sign between the terms. The common factor stays the same.

Worked example 4: retaining an addition sign

Question. Factorise fully $18a^2b+12ab^2$ .

Plan. Take out $6ab$ and keep the positive quotient from each term.

\begin{aligned} 18a^2b\div6ab&=3a &&\text{first quotient}\\ 12ab^2\div6ab&=2b &&\text{second quotient}\\ 18a^2b+12ab^2&=6ab(3a+2b) &&\text{write a product} \end{aligned}

Why this step? Factorising changes how the terms are grouped; it does not change their signs without a matching negative factor.

Answer. $6ab(3a+2b)$ .

Check. At $a=2$ , $b=1$ , the sum is $72+24=96$ , and the product is $12(8)=96$ .

The quotient calculation finds the factors, but the final polynomial identity also holds when $a=0$ or $b=0$ . We have not added a denominator to the answer. You can also take out a negative common factor when it makes the bracket easier to read.

Watch out

The line $18a^2b+12ab^2=3a(6ab+4b^2)$ is correct but is not fully factorised: $2b$ remains common inside. Taking it out gives $6ab(3a+2b)$ .

When Pearson asks “Factorise”, your answer must be a product. “Factorise fully” also asks you to check that no common factor remains.

Products of two linear factors

Each term in one bracket must multiply every term in the other. With two terms in each bracket, that gives four products. A linear expression, such as $x+4$ or $x-3$ , has highest power 1.

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Number

Algebra

Shape, space and measure

Handling data

2.2 - Expanding brackets and common factors

Distributing a single term

Taking out common factors

Products of two linear factors

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