Topics › KS3 › Algebra

Factorising

fac-01 Present from here

Factorise fully $6x + 9$

Option A: $3(2x + 9)$

Option B: $x(6 + 9)$

Option C: $3(2x + 3)$

Option D: $3(3x + 6)$

Answer and misconceptions

Option A: $3(2x + 9)$ ✗
Divides only the first term by the factorTakes 3 out of 6x but leaves 9 unchanged inside the bracket.

Option B: $x(6 + 9)$ ✗
Thinks the letter is a common factor of both termsTakes x outside even though 9 has no x.

Option C: $3(2x + 3)$ ✓ Correct

Option D: $3(3x + 6)$ ✗
Subtracts the factor instead of dividingWorks out 6 − 3 and 9 − 3 for the terms in the bracket.
fac-02 Present from here

Factorise fully $8a - 12$

Option A: $4(2a - 3)$

Option B: $2(4a - 6)$

Option C: $4(2a + 3)$

Option D: $4(2a - 12)$

Answer and misconceptions

Option A: $4(2a - 3)$ ✓ Correct

Option B: $2(4a - 6)$ ✗
Does not use the highest common factorTakes out 2, which is a common factor but not the highest, so it is not fully factorised.

Option C: $4(2a + 3)$ ✗
Loses the negative sign when factorisingDivides correctly but changes −12 to +3 inside the bracket.

Option D: $4(2a - 12)$ ✗
Divides only the first term by the factorTakes 4 out of 8a but leaves the 12 unchanged.
fac-03 Present from here

Factorise fully $x^2 + 5x$

Option A: $x(x + 5x)$

Option B: $x(x + 5)$

Option C: $5(x^2 + x)$

Option D: $x(x^2 + 5)$

Answer and misconceptions

Option A: $x(x + 5x)$ ✗
Does not divide the second term by xTakes x outside but leaves 5x unchanged in the bracket.

Option B: $x(x + 5)$ ✓ Correct

Option C: $5(x^2 + x)$ ✗
Thinks the number is a common factor of both termsTakes out 5 even though x² has no factor of 5.

Option D: $x(x^2 + 5)$ ✗
Does not divide the first term by xTakes x outside but leaves x² unchanged in the bracket.
fac-04 Present from here

Factorise $x^2 + 7x + 12$

Option A: $(x + 7)(x + 12)$

Option B: $(x + 6)(x + 2)$

Option C: $(x + 5)(x + 2)$

Option D: $(x + 3)(x + 4)$

Answer and misconceptions

Option A: $(x + 7)(x + 12)$ ✗
Copies the coefficients into the bracketsPuts 7 and 12 straight into the brackets without finding a factor pair.

Option B: $(x + 6)(x + 2)$ ✗
Checks the product but not the sumUses a factor pair of 12 without checking it adds to 7.

Option C: $(x + 5)(x + 2)$ ✗
Checks the sum but not the productFinds two numbers that add to 7 without checking they multiply to 12.

Option D: $(x + 3)(x + 4)$ ✓ Correct
fac-05 Present from here

Factorise $x^2 - 5x + 6$

Option A: $(x - 2)(x - 3)$

Option B: $(x + 2)(x + 3)$

Option C: $(x + 2)(x - 3)$

Option D: $(x - 6)(x + 1)$

Answer and misconceptions

Option A: $(x - 2)(x - 3)$ ✓ Correct

Option B: $(x + 2)(x + 3)$ ✗
Ignores the signs in the expressionFinds 2 and 3 but makes both positive, which gives +5x.

Option C: $(x + 2)(x - 3)$ ✗
Thinks a minus sign means one number is negativeSees −5x and makes just one number negative, which gives −6 not +6.

Option D: $(x - 6)(x + 1)$ ✗
Matches the sum but not the productFinds −6 and 1, which add to −5 but multiply to −6.