Factorising

Algebra

Questions marked Higher are aimed at the Higher tier.

These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.

Factorise $6x + 9$.

2 marks
Factorise $x^{2} - 16$.

2 marks
Factorise $x^{2} + 7x + 12$.

3 marks
Factorise $x^{2} - 2x - 15$.

3 marks
Factorise $2x^{2} + 7x + 3$.

4 marks
Factorise fully $3x^{2} - 12$.

4 marks
Factorise $6x^{2} - 7x - 3$. Higher

5 marks
Factorise $x^{2} - 9x + 18$ and hence solve $x^{2} - 9x + 18 = 0$. Then factorise $x^{3} - 9x^{2} + 18x$ and solve $x^{3} - 9x^{2} + 18x = 0$. Higher

6 marks

Mark scheme

Answer. $3(2x + 3)$
- The common factor is $3$.
- $3(2x + 3)$
Answer. $(x - 4)(x + 4)$
- Difference of squares.
- $(x - 4)(x + 4)$
Answer. $(x + 3)(x + 4)$
- Numbers that multiply to $12$ and add to $7$ are $3$ and $4$.
- $(x + 3)(x + 4)$
Answer. $(x - 5)(x + 3)$
- Numbers that multiply to $-15$ and add to $-2$ are $-5$ and $3$.
- $(x - 5)(x + 3)$
Answer. $(2x + 1)(x + 3)$
- Find a pair that multiplies to $6$ and adds to $7$.
- $(2x + 1)(x + 3)$
Answer. $3(x - 2)(x + 2)$
- $3(x^{2} - 4)$
- $3(x - 2)(x + 2)$
Answer. $(3x + 1)(2x - 3)$
- Look for two brackets whose product starts with $6x^{2}$ and ends with $-3$.
- $(3x + 1)(2x - 3) = 6x^{2} - 7x - 3$.
Answer. $(x - 6)(x - 3)$; $x = 6$ or $x = 3$; $x(x - 6)(x - 3)$; $x = 0$, $3$ or $6$
- $x^{2} - 9x + 18 = (x - 6)(x - 3)$.
- Solutions $x = 6$ and $x = 3$.
- $x^{3} - 9x^{2} + 18x = x(x - 6)(x - 3)$.
- Solutions $x = 0$, $x = 3$ and $x = 6$.