A4 - Three brackets and harder factorising (Higher) | AQA GCSE Mathematics

Longer expressions become manageable when you keep one stage visible at a time. Expand three brackets, factorise quadratics whose leading coefficient is not one, and work exactly with surd coefficients. The same distributive rule still does the work; the extra challenge is choosing a useful order.

Expanding three brackets (Higher)

Could you work out $999 \times 1000 \times 1001$ without long multiplication? Three consecutive factors hide a short algebraic calculation. We will return to it once we can move confidently between products and expanded expressions.

You already know

Two binomials produce four products before collecting. A difference of squares gives $(A - B)(A + B) = A^2 - B^2$ . Keep every term's sign when expanding.

To multiply three brackets, expand two first. Keep the third bracket untouched until you have collected the first result. Then multiply every term of that quadratic by every term of the remaining binomial. There are six products before collecting at this stage.

The order of multiplication can be chosen. A pair such as $(x - 2)(x + 2)$ is particularly useful because it produces only $x^2 - 4$ . This staged expansion method is not on the formulae sheet: learn it.

Worked example 1: three positive constants

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

Plan. Expand the last two brackets, then multiply their quadratic by $x + 1$ .

\begin{aligned} (x + 2)(x + 3) &= x^2 + 5x + 6 && \text{expand two first} \\ (x + 1)(x^2 + 5x + 6) &= x(x^2 + 5x + 6) + (x^2 + 5x + 6) && \text{distribute the first bracket} \\ &= x^3 + 5x^2 + 6x + x^2 + 5x + 6 && \text{six products} \\ &= x^3 + 6x^2 + 11x + 6 && \text{collect matching powers} \end{aligned}

Why this step? Collecting the first pair before using the third bracket keeps the work shorter and makes missing products easier to spot.

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

Check. At $x = 1$ , the product is $2 \times 3 \times 4 = 24$ , and the cubic is $1 + 6 + 11 + 6 = 24$ .

The next example changes only the sign of $1$ in the first bracket. The quadratic from the other two brackets stays the same.

Worked example 2: subtracting a whole quadratic

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

Plan. Reuse $(x + 2)(x + 3) = x^2 + 5x + 6$ , then multiply it by $x - 1$ .

\begin{aligned} (x - 1)(x^2 + 5x + 6) &= x(x^2 + 5x + 6) - (x^2 + 5x + 6) && \text{distribute} \\ &= x^3 + 5x^2 + 6x - x^2 - 5x - 6 && \text{subtract all three terms} \\ &= x^3 + 4x^2 + x - 6 && \text{collect matching powers} \end{aligned}

Why this step? The $-1$ multiplies the whole quadratic, so all three of its terms change sign.

Answer. $x^3 + 4x^2 + x - 6$ .

Check. At $x = 1$ , one original bracket is zero. The cubic also gives $1 + 4 + 1 - 6 = 0$ .

The same method handles non-unit coefficients. Choosing the cancelling pair first gives $(2x + 1)(x - 2)(x + 2) = (2x + 1)(x^2 - 4) = 2x^3 + x^2 - 8x - 4$ . Four brackets can be multiplied as two pairs, then their results multiplied; every term must still meet every term.

Watch out

Wrong: $x(x^2 + 5x + 6) = x^3 + 5x + 6$ . The outside $x$ multiplies every term. Right: $x^3 + 5x^2 + 6x$ .

AQA accepts a grid or clearly written products. Keep the intermediate quadratic visible so a later slip does not hide correct method.

Factorising a quadratic with a leading coefficient (Higher)

When the coefficient of $x^2$ is not one, the bracket entries must account for that coefficient as well as the constant and middle term.

For example, $(2x + 1)(3x + 2)$ expands to $6x^2 + 4x + 3x + 2 = 6x^2 + 7x + 2$ . The middle term is made from two cross-products. Splitting that middle term lets you recover two brackets by taking out common factors.

Splitting the middle term

For a quadratic $ax^2 + bx + c$ , with $a \ne 0$ , find numbers $r$ and $s$ such that

r + s = b \qquad\text{and}\qquad rs = ac

Write $bx = rx + sx$ , then factorise the two groups. When an integer factorisation exists, suitable integer $r$ and $s$ produce a common bracket.

Not on the formulae sheet: learn it.

Why use $ac$ ? If the brackets are $(mx + n)(px + q)$ , their cross-term coefficients are $mq$ and $np$ . Their product is $mqnp = (mp)(nq) = ac$ , while their sum is $b$ . That is exactly what the split is looking for.

Worked example 3: a positive middle coefficient

Question. Factorise $6x^2 + 11x + 3$ . Non-calculator.

Plan. Find numbers with product $6 \times 3 = 18$ and sum $11$ , then group.

\begin{aligned} 9 \times 2 &= 18 \qquad 9 + 2 = 11 && \text{choose the split} \\ 6x^2 + 11x + 3 &= 6x^2 + 9x + 2x + 3 && \text{split the middle term} \\ &= 3x(2x + 3) + 1(2x + 3) && \text{factorise each pair} \\ &= (3x + 1)(2x + 3) && \text{take out the common bracket} \end{aligned}

Why this step? The same bracket $2x + 3$ appears in both groups, so it is a common factor of the whole expression.

Answer. $(3x + 1)(2x + 3)$ .

Check. Expanding gives $6x^2 + 9x + 2x + 3 = 6x^2 + 11x + 3$ .

The next example changes only the sign of the middle coefficient. The split numbers must now both be negative.

Worked example 4: taking out a negative factor

Question. Factorise $6x^2 - 11x + 3$ . Non-calculator.

Plan. Split $-11x$ into $-9x - 2x$ , then choose factors that make the brackets match.

\begin{aligned} (-9)(-2) &= 18 \qquad -9 - 2 = -11 && \text{check product and sum} \\ 6x^2 - 11x + 3 &= 6x^2 - 9x - 2x + 3 && \text{split the middle term} \\ &= 3x(2x - 3) - 1(2x - 3) && \text{take out a negative }1 \\ &= (3x - 1)(2x - 3) && \text{common bracket} \end{aligned}

Why this step? Factoring $-1$ from $-2x + 3$ gives $-(2x - 3)$ . Both signs inside are needed to recover the original terms.

Answer. $(3x - 1)(2x - 3)$ .

Check. The cross-products are $-9x$ and $-2x$ , and the constants multiply to $+3$ .

A negative constant makes the split numbers have opposite signs. For example,

\begin{aligned} 6x^2 + x - 2 &= 6x^2 + 4x - 3x - 2 && 4(-3)=-12,\quad 4-3=1 \\ &= 2x(3x + 2) - 1(3x + 2) && \text{factorise each pair} \\ &= (2x - 1)(3x + 2) && \text{common bracket} \end{aligned}

Always look for a common factor first. For instance, $12x^2 + 22x + 6 = 2(6x^2 + 11x + 3) = 2(3x + 1)(2x + 3)$ . If no suitable integer pair exists, this method will not produce integer brackets; do not change the quadratic to force a pair.

Watch out

Wrong: for $6x^2 + 11x + 3$ , look for factors of $3$ with sum $11$ . That rule assumes the first coefficient is $1$ . Right: look for factors of $6 \times 3 = 18$ , with sum $11$ .

You can also use trial brackets, checking the leading product, constant product and both cross-products. The following quadratic has a negative constant, so the bracket constants have opposite signs.

Expanding expressions with surds (Higher)

A surd coefficient follows the same multiplication rules as an integer coefficient. Keep it exact, then collect only matching surd terms.

You already know

For non-negative $a$ and $b$ , $\sqrt{a}\sqrt{b} = \sqrt{ab}$ . In particular, $(\sqrt{a})^2 = a$ . Simplify roots such as $\sqrt{12} = 2\sqrt{3}$ before deciding which terms are alike.

For example, $\sqrt{2}\,x + 3\sqrt{2}\,x = 4\sqrt{2}\,x$ : each term contains the same variable and the same root. But $\sqrt{2}\,x + \sqrt{3}\,x$ does not combine into one multiple of a single one of those roots. The root product rule is not on the formulae sheet: learn it.

Worked example 5: matching square-root coefficients

Question. Expand and simplify $(x + \sqrt{2})(x + 3\sqrt{2})$ . Non-calculator.

Plan. Make the four products, square the repeated root, then collect the $x$ terms.

\begin{aligned} (x + \sqrt{2})(x + 3\sqrt{2}) &= x^2 + 3\sqrt{2}\,x + \sqrt{2}\,x + 3(\sqrt{2})^2 && \text{four products} \\ &= x^2 + 4\sqrt{2}\,x + 6 && \text{collect and square the root} \end{aligned}

Why this step? The constant product is $3 \times 2$ , while the middle terms are multiples of $\sqrt{2}\,x$ . They simplify in different ways.

Answer. $x^2 + 4\sqrt{2}\,x + 6$ .

Check. At $x = 0$ , the product is $\sqrt{2} \times 3\sqrt{2} = 6$ , the constant term in the answer.

The next example changes only the sign before the first root. The constant and one cross-product change sign.

Worked example 6: a negative surd term

Question. Expand and simplify $(x - \sqrt{2})(x + 3\sqrt{2})$ . Non-calculator.

Plan. Keep $-\sqrt{2}$ as a signed term throughout the expansion.

\begin{aligned} (x - \sqrt{2})(x + 3\sqrt{2}) &= x^2 + 3\sqrt{2}\,x - \sqrt{2}\,x - 3(\sqrt{2})^2 && \text{four signed products} \\ &= x^2 + 2\sqrt{2}\,x - 6 && \text{collect and simplify} \end{aligned}

Why this step? Only one factor of the constant product is negative, so that product is $-6$ .

Answer. $x^2 + 2\sqrt{2}\,x - 6$ .

Check. At $x = \sqrt{2}$ , the original has a zero bracket; the answer gives $2 + 4 - 6 = 0$ .

Different roots can occur too. Work exactly through the products:

\begin{aligned} (\sqrt{3} + \sqrt{2})(\sqrt{3} - 2\sqrt{2}) &= 3 - 2\sqrt{6} + \sqrt{6} - 4 && \text{multiply the roots} \\ &= -1 - \sqrt{6} && \text{collect matching terms} \end{aligned}

The coefficients $-2$ and $1$ combine because both multiply $\sqrt{6}$ . Neither root needs a decimal approximation.

Watch out

Wrong: $\sqrt{2} + \sqrt{3} = \sqrt{5}$ . The product rule applies to multiplication, not addition. Keep unlike surds separate; $\sqrt{2}\sqrt{3} = \sqrt{6}$ .

The next product keeps opposite signs and changes the root and its coefficient.

Recognising exact surd factors (Higher)

A quadratic with surd coefficients can sometimes be made familiar by noticing a repeated algebraic piece.

In $2x^2 + 3\sqrt{2}\,x + 2$ , the first term is $(\sqrt{2}\,x)^2$ , and the middle term is three times $\sqrt{2}\,x$ . If you temporarily call $\sqrt{2}\,x$ by the name $y$ , the quadratic becomes $y^2 + 3y + 2$ . That has the familiar factors $(y + 1)(y + 2)$ . Replace $y$ at the end.

This use of substitution changes the notation while keeping the value of the whole expression. It is not a new formula, and it is not on the formulae sheet: learn the strategy.

Worked example 7: a repeated surd factor

Question. Factorise $2x^2 + 3\sqrt{2}\,x + 2$ . Non-calculator.

Plan. Write $y = \sqrt{2}\,x$ , factorise in $y$ , then replace $y$ .

\begin{aligned} 2x^2 + 3\sqrt{2}\,x + 2 &= y^2 + 3y + 2 && y = \sqrt{2}\,x \\ &= (y + 1)(y + 2) && 1+2=3,\quad 1\times2=2 \\ &= (\sqrt{2}\,x + 1)(\sqrt{2}\,x + 2) && \text{replace }y \end{aligned}

Why this step? Both non-constant terms can be expressed using the same piece $\sqrt{2}\,x$ , which reduces the factorisation to one you already know.

Answer. $(\sqrt{2}\,x + 1)(\sqrt{2}\,x + 2)$ .

Check. Expanding gives $2x^2 + 2\sqrt{2}\,x + \sqrt{2}\,x + 2$ , the original quadratic.

The next example changes only the sign of the middle coefficient. Both bracket constants must now be negative.

Worked example 8: a negative middle surd coefficient

Question. Factorise $2x^2 - 3\sqrt{2}\,x + 2$ . Non-calculator.

Plan. Use $y = \sqrt{2}\,x$ again, then find constants with sum $-3$ and product $2$ .

\begin{aligned} 2x^2 - 3\sqrt{2}\,x + 2 &= y^2 - 3y + 2 && \text{use the same repeated piece} \\ &= (y - 1)(y - 2) && -1-2=-3,\quad (-1)(-2)=2 \\ &= (\sqrt{2}\,x - 1)(\sqrt{2}\,x - 2) && \text{replace }y \end{aligned}

Why this step? A positive constant product and negative sum require two negative constants, just as they did without surds.

Answer. $(\sqrt{2}\,x - 1)(\sqrt{2}\,x - 2)$ .

Check. The cross-products are $-2\sqrt{2}\,x$ and $-\sqrt{2}\,x$ , adding to $-3\sqrt{2}\,x$ .

Common factors still come first. For example, $2\sqrt{3}\,x^2 + 6x = 2\sqrt{3}\,x(x + \sqrt{3})$ . Expanding checks the second term: $2\sqrt{3}\,x \times \sqrt{3} = 6x$ . The general aim has not changed: the product of your factors must recover every term.

Watch out

Wrong: after using $y = \sqrt{2}\,x$ , leave the answer as $(y - 1)(y - 2)$ . The question is in $x$ . Right: replace every $y$ with $\sqrt{2}\,x$ in the final answer.

Now the opening product needs only the difference-of-squares pair:

\begin{aligned} 999 \times 1000 \times 1001 &= 1000(1000 - 1)(1000 + 1) && \text{choose the useful pair} \\ &= 1000(1000^2 - 1) && \text{difference of squares} \\ &= 1000^3 - 1000 && \text{expand one bracket} \\ &= 999\,999\,000 && \text{subtract }1000 \end{aligned}

That is the satisfying part of algebra: a large calculation can become a small one when you choose the form well.

Quick Recap

Expand three brackets in stages. Look first for a pair whose cross-products cancel.
For $ax^2 + bx + c$ , split the middle term using numbers with product $ac$ and sum $b$ , then group.
Surd coefficients follow the same product rules. Keep roots exact and collect only matching terms.
A repeated piece such as $\sqrt{2}\,x$ can be given a temporary name to reveal familiar factors. Replace it before finishing.

The opening product is $999\,999\,000$ : using $(1000 - 1)(1000 + 1)$ saved the long multiplication.

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A4 - Three brackets and harder factorising (Higher)

Expanding three brackets (Higher)

Factorising a quadratic with a leading coefficient (Higher)

Expanding expressions with surds (Higher)

Recognising exact surd factors (Higher)

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