E2.2 - Several brackets and grouped factors | Cambridge IGCSE Mathematics

This Extended lesson takes the distributive law beyond two brackets in one variable. You will expand products with several letters or three brackets, then reverse expansion to factorise grouped terms and cubic expressions with a common factor.

Products with several variables

How can four separate terms become a product of two brackets? To reverse a multiplication confidently, first make the forward multiplication visible. With two different letters, every term still multiplies every term, but not every resulting term has the same letter part.

For instance, $(a+b)(a+2b)$ produces $a^2$ , $2ab$ , $ab$ and $2b^2$ . Only the two $ab$ -terms can be collected. At $a=3$ , $b=1$ , the product is $4\times5=20$ , agreeing with $9+6+3+2=20$ .

You already know

Letter order in a product does not change its value: $ab=ba$ . Powers do matter: $a^2$ , $ab$ and $b^2$ are different letter parts.

A product grid is especially helpful when the brackets contain different variables. Put one complete term in each row or column header, including its sign.

Worked example 1: two letters in both brackets

Question. Expand and simplify $(2x+y)(x+3y)$ .

Plan. Form four products and combine only the two $xy$ -terms.

\begin{aligned} (2x+y)(x+3y)&=(2x)(x)+(2x)(3y)+y(x)+y(3y) && \text{all term pairings}\\ &=2x^2+6xy+xy+3y^2 && \text{write }yx=xy\\ &=2x^2+7xy+3y^2 && \text{collect like terms} \end{aligned}

Answer. $2x^2+7xy+3y^2$ .

Check. At $x=2$ , $y=1$ , the brackets give $5\times5=25$ . The expansion gives $8+14+3=25$ .

Change only the sign of $y$ in the first bracket. Its two products change sign; the two products from $2x$ do not.

Worked example 2: changing one signed term

Question. Expand and simplify $(2x-y)(x+3y)$ .

Plan. Multiply the term $-y$ by each term in the second bracket.

\begin{aligned} (2x-y)(x+3y)&=(2x)(x)+(2x)(3y)+(-y)(x)+(-y)(3y) && \text{retain the minus sign}\\ &=2x^2+6xy-xy-3y^2 && \text{multiply}\\ &=2x^2+5xy-3y^2 && \text{collect the }xy\text{-terms} \end{aligned}

Answer. $2x^2+5xy-3y^2$ .

Check. At $x=2$ , $y=1$ , the brackets give $3\times5=15$ . The expansion gives $8+10-3=15$ .

If a product contains powers, keep the complete letter part. For example, $(x^2)(y)=x^2y$ , while $x(y^2)=xy^2$ ; these cannot be collected.

Watch out

$6xy+xy=7x^2y^2$ is wrong. Addition combines the coefficients, giving $7xy$ . Squaring the letters would change the quantities being added.

One substitution can check your arithmetic; the distributive steps justify the expansion generally.

Expanding three brackets

With three brackets, multiply two first and treat their answer as one expression. Then multiply that entire expression by the remaining bracket. Grouping the factors in this way is valid because $(AB)C=A(BC)$ .

Choose a convenient pair; the final answer is independent of the choice. Keep the unused bracket on your working line so it is not lost. A quadratic expression times a two-term bracket produces six products before collecting.

Worked example 3: three positive brackets

Question. Expand and simplify $(x+1)(x+2)(x+3)$ .

Plan. Expand the first two brackets, then distribute all three resulting terms over the last bracket.

\begin{aligned} (x+1)(x+2)&=x^2+3x+2 && \text{expand and collect}\\ (x^2+3x+2)(x+3) &=x^3+3x^2+3x^2+9x+2x+6 && \text{three terms times two}\\ &=x^3+6x^2+11x+6 && \text{collect each power} \end{aligned}

Here $x^2\times x=x^3$ because there are three factors of $x$ . An expression whose highest power is $x^3$ , with nonzero coefficient, is called cubic.

Answer. $x^3+6x^2+11x+6$ .

Check. At $x=2$ , the original is $3\times4\times5=60$ ; the expansion is $8+24+22+6=60$ .

Change only the final $+3$ to $-3$ . The first two-bracket result is unchanged, but some later terms cancel.

Worked example 4: cancellation after a third bracket

Question. Expand and simplify $(x+1)(x+2)(x-3)$ .

Plan. Reuse the first pair's expansion and multiply it by $x-3$ .

\begin{aligned} (x+1)(x+2)(x-3)&=(x^2+3x+2)(x-3) && \text{expand the first pair}\\ &=x^3-3x^2+3x^2-9x+2x-6 && \text{include all six products}\\ &=x^3-7x-6 && \text{the }x^2\text{-terms cancel} \end{aligned}

Answer. $x^3-7x-6$ .

Check. At $x=2$ , the brackets give $3\times4\times(-1)=-12$ . The answer gives $8-14-6=-12$ .

The missing $x^2$ -term in the final answer means its coefficient is zero; it does not mean the expression has stopped being cubic. The highest nonzero power is still $x^3$ .

Watch out

Multiplying $x^2+3x+2$ by $x-3$ does not mean multiplying only $x^2$ by the last bracket. Each of the three terms must multiply both $x$ and $-3$ .

Writing the six-product line makes cancellation visible before you simplify.

Factorising by grouping and common factors

Sometimes no letter is shared by every term, but a whole bracket appears after you group the terms in pairs. For example, $ax+bx=x(a+b)$ , while $3ay+3by=3y(a+b)$ . The shared factor is now the entire bracket $a+b$ .

Factorisation by grouping

Factorisation by grouping rewrites groups of terms to reveal a common factor, then takes out that factor. The common factor can be an expression in brackets.

This is the same distributive law as before: $xA+3yA=A(x+3y)$ . Here the single placeholder $A$ stands for the whole expression $a+b$ . Put it back after the structure is clear.

Worked example 5: a common bracket

Question. Factorise $ax+bx+3ay+3by$ .

Plan. Group the first pair and the last pair, then take out their common bracket.

\begin{aligned} ax+bx+3ay+3by&=x(a+b)+3y(a+b) && \text{factorise each pair}\\ &=(a+b)(x+3y) && \text{take out }a+b \end{aligned}

Answer. $(a+b)(x+3y)$ .

Check. Expanding gives $ax+3ay+bx+3by$ , exactly the original terms in a different order.

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E2.2 - Several brackets and grouped factors

Products with several variables

Expanding three brackets

Factorising by grouping and common factors

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