Lesson 3 ~40 min Algebraic Techniques C · Path +85 XP

Factorising: Common Factors, Two Squares and Grouping

Factorising is expanding run backwards, and the hardest part is knowing which method to reach for. Three strategies handle most expressions, and they have a reliable order.

Today's hook: Factorise $6x^3 - 24x$. Reach straight for the difference of two squares and you will get stuck, because it is not one. Take out the common factor first and it becomes $6x(x^2 - 4)$, and now the difference of two squares finishes the job. Order matters.

0/5QUESTS

Think First

warm-up

Expand $(x + 7)(x - 7)$ and write down the result. Now cover your working and look only at the answer. What features of it would tell someone, who had not seen the question, that it came from a pair of brackets like that? List every clue you can find.

Record your answer in your workbook.

The Big Idea

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To factorise is to write an expression as a product. Every method is a pattern being recognised in reverse, so the skill is diagnosis before technique. Always take out the highest common factor first: it is never wrong, it often exposes a second pattern underneath, and forgetting it is the most common way a correct method still gives an incomplete answer.

$$\text{HCF first} \;\longrightarrow\; \text{count the terms} \;\longrightarrow\; \text{choose the pattern}$$

After the common factor comes out, count the terms. Two terms suggest a difference of two squares. Four terms suggest grouping in pairs. Three terms is a quadratic trinomial, which is Lesson 4. Counting is a better first question than staring.

$a^2 - b^2 = (a+b)(a-b)$

HCF always first

It costs nothing, and it often turns a hard expression into an easy one.

Two terms, two squares

Only a DIFFERENCE factorises. $a^2 + b^2$ does not.

Check by expanding

Multiply your factors back out. It should return the original exactly.

What You'll Master

objectives

Know

That the highest common factor is always removed first
That $a^2 - b^2 = (a+b)(a-b)$, and that a SUM of two squares does not factorise
That a four-term expression is a candidate for grouping in pairs

Understand

Why removing the common factor can expose a further factorisation
Why grouping works: the pairs must leave the SAME bracket behind

Can Do

Factorise by taking out the highest common factor, including negative and algebraic factors
Factorise a difference of two squares, including after a common factor is removed
Factorise a four-term expression by grouping in pairs, choosing the pairing that works

Words You Need

vocabulary

FactoriseTo write an expression as a product of factors.

Highest common factorThe largest factor shared by every term, including any common letters.

Difference of two squares$a^2 - b^2$, which factorises as $(a+b)(a-b)$.

Grouping in pairsSplitting a four-term expression into two pairs, each with its own common factor.

Fully factorisedFactorised until no factor can be broken down further.

Perfect square numberA number that is the square of a whole number, such as $49$ or $121$.

The Highest Common Factor, First and Always

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The highest common factor takes both the numbers and the letters into account. For $12x^3 - 18x^2$, the numbers share $6$ and the letters share $x^2$, so the HCF is $6x^2$:

$$12x^3 - 18x^2 = 6x^2(2x - 3)$$

Check by expanding: $6x^2 \times 2x = 12x^3$ and $6x^2 \times (-3) = -18x^2$. Correct.

When the leading term is negative it is usually tidier to take the negative out as well:

$$-10x^2 - 15x = -5x(2x + 3)$$

Both signs inside the bracket flip when the negative comes out, which is worth checking every time.

Why first

$3x^2 - 27$ is not a difference of two squares, because $3x^2$ is not a perfect square. Remove the $3$ and it becomes $3(x^2 - 9)$, which is.

The Difference of Two Squares

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Two terms, both perfect squares, separated by a minus. That is the whole diagnosis.

$$x^2 - 49 = (x)^2 - (7)^2 = (x + 7)(x - 7)$$

The squares can carry coefficients, and each part must be square-rooted whole:

$$25m^2 - 64n^2 = (5m)^2 - (8n)^2 = (5m + 8n)(5m - 8n)$$

A sum does not factorise. There is no pair of brackets with real numbers whose product is $x^2 + 9$. Expanding any $(x+p)(x+q)$ gives a middle term of $(p+q)x$, and making that zero forces $q = -p$, which turns the constant into $-p^2$, a negative. So the constant can never come out positive.

Combined with a common factor, the pattern often appears one step in:

$$2y^3 - 50y = 2y(y^2 - 25) = 2y(y+5)(y-5)$$

Grouping in Pairs

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Four terms and no common factor across all of them is the signal to group. Split into two pairs, factorise each pair, and look for a shared bracket.

$$xy + 3x + 2y + 6$$

Pair the first two and the last two. The first pair shares $x$; the second shares $2$:

$$= x(y + 3) + 2(y + 3)$$

Both pairs left behind the same bracket, $(y+3)$. That bracket is now itself a common factor:

$$= (y + 3)(x + 2)$$

If the two brackets do not match, the pairing was wrong, not the method. Try pairing the first and third terms instead. And when the third term is negative, take out a NEGATIVE factor so the bracket still matches:

$$ab + 4a - 3b - 12 = a(b+4) - 3(b+4) = (b+4)(a-3)$$

Choosing a Strategy

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Work down this list. The first question is always the same one.

Step	Ask	If yes
1	Is there a common factor?	Take it out, then start again on what is left
2	Two terms, both squares, subtracted?	Difference of two squares
3	Four terms?	Group in pairs
4	Three terms?	Quadratic trinomial, see Lesson 4

Step 1 is not optional and it is not last. Applying it to $6x^3 - 24x$ gives $6x(x^2 - 4)$, and step 2 then finishes: $6x(x+2)(x-2)$. Skipping step 1 leaves you staring at an expression that matches no pattern at all.

Common Pitfalls

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Stopping after one factorisation: leaving $3(x^2 - 9)$ as the final answer.

Fix: after every step, look at each factor again. $x^2 - 9$ is a difference of two squares, so the answer is $3(x+3)(x-3)$.

Trying to factorise a SUM of two squares, such as $x^2 + 16$.

Fix: only a difference factorises. $x^2 + 16$ has no factorisation over the real numbers, and saying so is the correct answer.

Grouping and getting two different brackets, then concluding the expression will not factorise.

Fix: try a different pairing first. $xy + 2y + 3x + 6$ fails if you pair badly but works as $y(x+2) + 3(x+2)$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Common factor, then a second pattern

+15 XP per step

PROBLEM

Factorise $18x^3 - 8x$ fully.

1

Look for the HCF

$\text{numbers share } 2, \quad \text{letters share } x$

Always step 1, before any pattern is considered.
2

Take it out

$18x^3 - 8x = 2x(9x^2 - 4)$

Check: $2x \times 9x^2 = 18x^3$ and $2x \times (-4) = -8x$.
3

Look again at what is left

$9x^2 - 4 = (3x)^2 - (2)^2$

Two terms, both perfect squares, subtracted.
4

Apply the difference of two squares

$2x(3x + 2)(3x - 2)$

No factor can be broken down further, so this is fully factorised.

Answer$2x(3x+2)(3x-2)$

Watch Me Solve It · Grouping with a negative

+15 XP per step

PROBLEM

Factorise $6ac + 9ad - 2bc - 3bd$.

1

Check for a common factor across all four

$\text{none: } 6, 9, 2, 3 \text{ share no factor}$

So move to the four-term strategy.
2

Pair the first two and the last two

$(6ac + 9ad) + (-2bc - 3bd)$

Keep the signs with their terms.
3

Factorise each pair, taking a NEGATIVE out of the second

$3a(2c + 3d) - b(2c + 3d)$

Taking $-b$ out makes the second bracket match the first; taking $+b$ would not.
4

Take out the shared bracket

$(2c + 3d)(3a - b)$

Check by expanding: the four products return the original.

Answer$(2c + 3d)(3a - b)$

Watch Me Solve It · Deciding what will not factorise

+15 XP per step

PROBLEM

Factorise fully where possible: (a) $4x^2 - 81$, (b) $4x^2 + 81$, (c) $5x^2 - 45$.

1

Part (a): check the pattern

$4x^2 - 81 = (2x)^2 - (9)^2 = (2x+9)(2x-9)$

Two terms, both squares, subtracted.
2

Part (b): check the sign

$4x^2 + 81 \text{ is a SUM of two squares}$

A sum has no real factorisation, and stating that is the complete answer.
3

Part (c): HCF first

$5x^2 - 45 = 5(x^2 - 9)$

$5x^2$ is not a perfect square, so the pattern does not apply until the 5 is removed.
4

Then the pattern

$5(x + 3)(x - 3)$

Answer(a) $(2x+9)(2x-9)$, (b) does not factorise, (c) $5(x+3)(x-3)$

Brain Trainer · 4 problems

Brain Trainer · Factorise fully

4 problems

Four quick problems. Work each one, then reveal the answer.

1 Factorise $15x^2 - 20x$.

HCF is $5x$.$5x(3x - 4)$
2 Factorise $m^2 - 144$.

Two terms, both squares, subtracted. $144 = 12^2$.$(m+12)(m-12)$
3 Factorise $3y^3 - 27y$ fully.

HCF $3y$ first, leaving $y^2 - 9$.$3y(y+3)(y-3)$
4 Factorise $pq - 5p + 2q - 10$.

Group: $p(q-5) + 2(q-5)$.$(q-5)(p+2)$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

The first step

+10 XP

When factorising any expression, the first thing to check for is:

MC2

Two squares

+10 XP

$49y^2 - 16$ factorises to:

MC3

A sum of squares

+10 XP

Which expression does NOT factorise over the real numbers?

MC4

Grouping

+10 XP

$ax + ay + bx + by$ factorises to:

MC5

Fully factorised

+10 XP

A student factorises $4x^3 - 36x$ as $4x(x^2 - 9)$. This answer is:

Short Answer · 3 questions

Factorise fully

+15 XP

SHORT ANSWER

Factorise each expression completely.
(a) $24a^3 b - 16a^2 b^2$
(b) $50 - 2t^2$
(c) $3mn - 12m + 5n - 20$
(d) $x^4 - 16$

Write your working in your book.

Diagnose and justify

+15 XP

SHORT ANSWER

For each expression, state which strategy applies and why, then factorise or explain why it cannot be factorised.
(a) $9p^2 + 25$
(b) $7x^2 - 63$
(c) $2ab - 6a + b - 3$

Write your working in your book.

Work backwards

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SHORT ANSWER

(a) An expression factorises to $5x(x + 4)(x - 4)$. Expand it fully to find the original expression.
(b) A four-term expression factorises to $(y - 2)(3y + 7)$... but a factorised four-term expression should give a product of two binomials. Expand $(y-2)(3y+7)$ and explain how many terms the ORIGINAL grouping expression would have had before collecting.
(c) Explain why checking a factorisation by expansion is a complete check, whereas substituting one value is not.

Write your working in your book.

Stretch Challenge · Hidden differences

+25 XP

CHALLENGE

(a) Factorise $x^4 - 81$ completely, and state which factor cannot be broken down further and why.
(b) Factorise $(x + 3)^2 - 16$ completely, treating $(x+3)$ as a single quantity.
(c) Hence factorise $(2m - 1)^2 - (m + 4)^2$ completely, and simplify each factor.

Quick Review

recap

Step 1

Highest common factor, every time

Two terms

$a^2 - b^2 = (a+b)(a-b)$; a sum does not

Four terms

Group in pairs; the brackets must match

Finished?

Re-examine every factor, then expand to check

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