Factorising into a single bracket probing questions

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Convince Me That…

Students must construct a mathematical argument for why each statement is true.

1

Convince me that $ 2(6x + 10) $ is not the fully factorised form of $ 12x + 20 $

💡 Possible Argument

While $ 2(6x + 10) $ does expand to give $ 12x + 20 $, it is not fully factorised because 2 is not the highest common factor of 12 and 20. The factors of 12 include 1, 2, 3, 4, 6, 12 and the factors of 20 include 1, 2, 4, 5, 10, 20. The HCF is 4, so the fully factorised form is $ 4(3x + 5) $. You can check that the terms inside the bracket, 3 and 5, share no common factor other than 1.

A quick way to spot that you haven’t finished factorising: look at the terms inside the bracket. $ 6x + 10 $ — do 6 and 10 share a common factor? Yes, 2. So there’s more factorising to do. The expression is only fully factorised when the terms inside have no common factor.

2

Convince me that $ 6x^2 + 10x $ can be factorised even though it has no constant term

💡 Possible Argument

Even though there is no constant term, both terms share the common factor $ 2x $. We have $ 6x^2 = 2x \times 3x $ and $ 10x = 2x \times 5 $, so $ 6x^2 + 10x = 2x(3x + 5) $. The absence of a constant term actually makes the variable $ x $ a common factor — every term contains at least one $ x $.

In fact, expressions like this are often easier to factorise than those with constant terms, because the shared $ x $ is visible in every term. Expanding confirms: $ 2x(3x + 5) = 2x \times 3x + 2x \times 5 = 6x^2 + 10x $ ✓.

3

Convince me that the factorised form of $ 7x + 7 $ is $ 7(x + 1) $, not just $ 7x $ or $ 7(x) $

💡 Possible Argument

When we factorise, we divide every term by the common factor. $ 7x \div 7 = x $, and $ 7 \div 7 = 1 $. The 1 must remain in the bracket to hold the place of the second term. If we just wrote $ 7(x) $, expanding it would give $ 7x $, not $ 7x + 7 $. We need the $ +1 $ so that $ 7 \times 1 = 7 $ appears in the answer.

Think of it as: “How many 7s are in 7x? (x). How many 7s are in 7? (1).”

4

Convince me that $ 4x \; – \; 12 $ factorises to $ 4(x \; – \; 3) $, not $ 4(x + 3) $

💡 Possible Argument

The original expression is $ 4x \; – \; 12 $. When we take out the factor of 4, we divide each term by 4: $ 4x \div 4 = x $ and $ -12 \div 4 = -3 $. The minus sign belongs to the 12 and must be carried into the bracket, giving $ 4(x \; – \; 3) $. We can check: $ 4(x \; – \; 3) = 4x \; – \; 12 $ ✓, but $ 4(x + 3) = 4x + 12 $ ✗ — the sign of the constant is wrong.

A common error is to ignore the sign when dividing the constant term by the common factor. Students should remember that the sign “stays with” the term: if the original has subtraction, the bracket will too (unless a negative factor is taken out).

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Give an Example Of…

Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.

1

Give an example of an expression with two terms that factorises with 6 outside the bracket

An example

Another example

One no-one else will think of

A sneaky non-example

💡 Possible Answers

Example: $ 6x + 18 = 6(x + 3) $

Another: $ 12y + 30 = 6(2y + 5) $

Creative: $ 54a \; – \; 6 = 6(9a \; – \; 1) $ — students rarely choose expressions where 6 is much smaller than one of the coefficients. Or $ 6x^2 + 6 = 6(x^2 + 1) $.

Trap: $ 6x + 9 $ — a student might see the 6 in the first term and assume 6 can be factored out. But 6 is not a factor of 9, so the HCF is actually 3, giving $ 3(2x + 3) $, not $ 6(\ldots) $.

2

Give an example of an expression that factorises with $ x $ as part of the common factor

An example

Another example

One no-one else will think of

A sneaky non-example

💡 Possible Answers

Example: $ x^2 + 5x = x(x + 5) $

Another: $ 3x^2 + 12x = 3x(x + 4) $

Creative: $ x^3 + x^2 = x^2(x + 1) $ — every term contains $ x^2 $, so the highest power of $ x $ common to all terms goes outside. Or $ 7x^2 \; – \; 21x = 7x(x \; – \; 3) $.

Trap: $ x^2 + 5 $ — a student might think the $ x^2 $ means $ x $ is a common factor. But 5 does not contain $ x $, so $ x $ is not a common factor of both terms. This expression cannot be factorised into a single bracket.

3

Give an example of an expression where taking out a factor of 2 does not give the fully factorised form

An example

Another example

One no-one else will think of

A sneaky non-example

💡 Possible Answers

Example: $ 12x + 8 $ — taking out 2 gives $ 2(6x + 4) $, but 6 and 4 still share a factor of 2, so the HCF is actually 4: $ 4(3x + 2) $.

Another: $ 6x + 18 $ — taking out 2 gives $ 2(3x + 9) $, but 3 and 9 share a factor of 3, so the HCF is 6: $ 6(x + 3) $.

Creative: $ 20x^2 + 30x $ — taking out 2 gives $ 2(10x^2 + 15x) $, but the HCF is $ 10x $: $ 10x(2x + 3) $. This combines both a larger numerical factor and a variable factor that the student missed.

Trap: $ 2x + 7 $ — a student might offer this, thinking 2 is not the full factorisation. But HCF(2, 7) = 1, so $ 2x + 7 $ cannot be factorised at all. Taking out 2 is not possible here without introducing fractions, so it doesn’t satisfy the condition.

4 ✦

Give an example of an expression where a negative number is the common factor taken outside the bracket

An example

Another example

One no-one else will think of

A sneaky non-example

💡 Possible Answers

Example: $ -3x \; – \; 12 = -3(x + 4) $ — check: $ -3 \times x + (-3) \times 4 = -3x \; – \; 12 $ ✓

Another: $ -5x + 20 = -5(x \; – \; 4) $ — check: $ -5 \times x + (-5) \times (-4) = -5x + 20 $ ✓

Creative: $ -2x^2 + 8x = -2x(x \; – \; 4) $ — both a negative and a variable in the common factor. Or $ -x \; – \; 1 = -(x + 1) $ — taking out −1 where the “invisible” coefficient is −1.

Trap: $ 3x \; – \; 12 = 3(x \; – \; 4) $ — a student might think the subtraction sign means the factor is negative. But the common factor is +3, not −3. The minus sign is part of the −12 term, not an indicator that the overall factor is negative.

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Always, Sometimes, Never

Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.

1

An expression can be fully factorised in more than one way, giving different results

NEVER

The fully factorised form of an expression is unique (assuming we are factoring out integers and keeping positive leading terms where possible). The HCF of the terms is a single, specific number. For example, the HCF of 12 and 18 is exactly 6, so $ 12x + 18 $ can only be fully factorised as $ 6(2x + 3) $.

Students sometimes confuse partial factorisations (like $ 2(6x + 9) $ or $ 3(4x + 6) $) with full factorisation. While there are multiple partial factorisations, there is exactly one full factorisation.

2

If you factorise an expression and then expand your answer, you get back to the original expression

ALWAYS

Factorising and expanding are inverse (reverse) operations — they undo each other. If $ 12x + 8 = 4(3x + 2) $, then expanding gives $ 4 \times 3x + 4 \times 2 = 12x + 8 $. This is always true because factorising is just rewriting an expression using the distributive law in reverse.

This is a useful classroom strategy: students can always check their factorisation by expanding the bracket and seeing if they get back to the original expression. If they don’t, they know something has gone wrong.

3

When you fully factorise an expression of the form $ ax + b $, the coefficient of $ x $ inside the bracket is 1

SOMETIMES

This is true when the HCF equals $ a $ — i.e., when $ a $ divides into $ b $. For example, $ 3x + 12 = 3(x + 4) $ because HCF(3, 12) = 3, and dividing $ 3x $ by 3 gives $ x $. But when the HCF is smaller than $ a $, the coefficient of $ x $ inside will be greater than 1. For example, $ 10x + 15 = 5(2x + 3) $ because HCF(10, 15) = 5, and dividing $ 10x $ by 5 gives $ 2x $.

True case: $ 5x + 20 = 5(x + 4) $ — coefficient of $ x $ inside is 1. False case: $ 10x + 15 = 5(2x + 3) $ — coefficient of $ x $ inside is 2.

4

The number outside the bracket is larger than every number inside the bracket

SOMETIMES

Consider $ 6x + 3 = 3(2x + 1) $ — the 3 outside is larger than both the 2 and the 1 inside, so the statement holds here. But $ 2x + 14 = 2(x + 7) $ — the 2 outside is smaller than the 7 inside, so it fails.

The number outside is the HCF. The numbers inside are what remains after division. Either could be larger.

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Odd One Out

Which is the odd one out? Can you make a case for each one? There’s no single right answer!

1

Which is the odd one out?

$ 6x + 12 $

$ 6x + 9 $

$ 6x + 4 $

💡 A Case for Each

$ 6x + 12 $ is the odd one out — it’s the only one that factorises with the full coefficient of $ x $ outside the bracket, giving $ 6(x + 2) $ where the $ x $ term inside has coefficient 1.

$ 6x + 9 $ is the odd one out — it’s the only one with an odd number as the HCF. (HCF is 3, whereas the others have even HCFs of 6 and 2).

$ 6x + 4 $ is the odd one out — it’s the only one with an HCF of 2. (The others have HCFs of 6 and 3, both of which are multiples of 3.)

2

Which is the odd one out?

$ 2(5x + 3) $

$ 5(2x + 3) $

$ 10(x + 3) $

💡 A Case for Each

$ 2(5x + 3) $ is the odd one out — it’s the only one where the coefficient of $ x $ inside the bracket is larger than the constant term inside ($5 > 3$). In the others, the constant is larger than the coefficient ($3 > 2$ and $3 > 1$).

$ 5(2x + 3) $ is the odd one out — it’s the only one with an odd number outside the bracket. (The other two have even numbers outside: 2 and 10.)

$ 10(x + 3) $ is the odd one out — it’s the only one where the coefficient of $ x $ inside the bracket is 1, meaning the full coefficient has been factored out.

3

Which is the odd one out?

$ x^2 + 5x $

$ 3x + 12 $

$ x^2 + 4 $

💡 A Case for Each

$ x^2 + 4 $ is the odd one out — it’s the only one that cannot be factorised into a single bracket, because the two terms share no common factor other than 1.

$ 3x + 12 $ is the odd one out — it’s the only one where the common factor is purely numerical (3), with no variable component.

$ x^2 + 5x $ is the odd one out — it’s the only one where a variable ($ x $) appears outside the bracket in the factorised form: $ x(x + 5) $.

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Explain the Mistake

Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?

1

Factorise $ 8x + 20 $

A student writes:

Answer: $ 4(2x + 20) $

Reasoning: “I found that 4 goes into 8, so I divided the 8x by 4 to get 2x. Then I kept the 20 as it is.”

🔍 The Mistake

The student has only divided the first term by 4 and left the second term unchanged. This is the “only factorise one term” misconception — the student treats factorising as dividing just the first coefficient. When you take out a common factor, you must divide every term by it. Since $ 20 \div 4 = 5 $, the correct answer is $ 4(2x + 5) $. Check: $ 4(2x + 5) = 8x + 20 $ ✓, but $ 4(2x + 20) = 8x + 80 $ ✗.

The student can verify by expanding: if the expanded form doesn’t match the original, they’ve made an error. This check catches the mistake immediately.

2

Factorise $ 30x + 45 $

A student writes:

Answer: $ 5(6x + 9) $

Reasoning: “30 and 45 both end in 5 or 0, so they’re both in the 5 times table. I divided both numbers by 5.”

🔍 The Mistake

The student’s logic is “correct” arithmetically (5 is a factor) but incorrect algebraically because it isn’t the Highest Common Factor. This leads to a partial factorisation. Look at the bracket: $ 6x + 9 $. Do 6 and 9 still share a factor? Yes, 3. This means we haven’t finished.

The HCF of 30 and 45 is actually 15. The fully factorised form is $ 15(2x + 3) $. The “divisibility trick” (ends in 5 or 0) often stops students looking for larger factors.

3

Factorise fully $ 12x + 18 $

A student writes:

Answer: $ 2(6x + 9) $

Reasoning: “12 and 18 are both even numbers, so I took out 2.”

🔍 The Mistake

The student has found a common factor (2), but not the highest common factor. This is the “not fully factorising” misconception. The expression inside the bracket, $ 6x + 9 $, can itself be factorised further since HCF(6, 9) = 3: $ 6x + 9 = 3(2x + 3) $. So the full factorisation is $ 6(2x + 3) $, where HCF(12, 18) = 6.

A useful strategy is to check after factorising: look at the terms inside the bracket and ask “do these still share a common factor?” If yes, you haven’t finished. The expression is only fully factorised when the terms inside the bracket are coprime (HCF = 1).

4

Factorise $ 6x \; – \; 15 $

A student writes:

Answer: $ 3(2x + 5) $

Reasoning: “3 goes into both 6 and 15. 6 divided by 3 is 2x and 15 divided by 3 is 5.”

🔍 The Mistake

The student correctly identified 3 as the HCF and correctly divided both coefficients, but lost the minus sign. This is the “dropping the sign” misconception.

Try the “Cover Up” method: Imagine covering up the $ 6x $ with your hand. You are left with $ -15 $. If you divide $ -15 $ by 3, you get $ -5 $, not $ +5 $. The sign belongs to the number. The correct answer is $ 3(2x \; – \; 5) $.

Factorising into a Single Bracket

Convince Me That…

Give an Example Of…

Always, Sometimes, Never

Odd One Out

Explain the Mistake