Factorise fully, including variable and negative common factors
7th grade (Year 8 in England) Β· 16 questions Β· Common Core: 7.EE.A.1 Β· England: KS3 Algebra, GCSE Foundation Algebra
Name
Date
Score/ 16
Question 1
Factorise β6x β 12 by taking out a negative common factor.
- A. β6(x + 2)
- B. β6(x + 12)
- C. β(6x β 12)
- D. 6(x + 2)
Question 2
Mateo factorises 12x + 4 as 2(6x + 2). Is it fully factorised?
- A. Yes
- B. No, it should be 4(3x + 1)
Question 3
Factorise fully: 6xy + 48x
- A. 6x(y + 8)
- B. 6x(y + 48)
- C. 6y(x + 8)
- D. 6(xy + 8x)
Question 4
Factorise β5x β 40 by taking out a negative common factor.
- A. β5(x + 40)
- B. β(5x β 40)
- C. β5(x + 8)
- D. 5(x + 8)
Question 5
Luis factorises 12x + 30 as 2(6x + 15). Is it fully factorised?
- A. Yes
- B. No, it should be 6(2x + 5)
Question 6
Factorise fully: 5xy + 10x
- A. 5x(y + 2)
- B. 5x(y + 10)
- C. 5y(x + 2)
- D. 5(xy + 2x)
Question 7
Factorise β4x β 16 by taking out a negative common factor.
- A. β(4x β 16)
- B. 4(x + 4)
- C. β4(x β 4)
- D. β4(x + 4)
Question 8
Luis factorises 18x + 24 as 6(3x + 4). Is it fully factorised?
- A. Yes
- B. No, it should be 6(3x + 4)
Question 9
Factorise fully: 2xy + 12x
- A. 2y(x + 6)
- B. 2(xy + 6x)
- C. 2x(y + 6)
- D. x(2y + 12)
Question 10
Factorise β3x β 24 by taking out a negative common factor.
- A. β3(x + 8)
- B. β(3x β 24)
- C. β3(x β 8)
- D. β3(x + 24)
Question 11
Hiro factorises 24x + 42 as 2(12x + 21). Is it fully factorised?
- A. Yes
- B. No, it should be 6(4x + 7)
Question 12
Factorise fully: 7xy + 21x
- A. 7y(x + 3)
- B. 7x(y + 3)
- C. 7(xy + 3x)
- D. x(7y + 21)
Question 13
Factorise β6x β 54 by taking out a negative common factor.
- A. β6(x β 9)
- B. 6(x + 9)
- C. β6(x + 54)
- D. β6(x + 9)
Question 14
Diego factorises 16x + 8 as 8(2x + 1). Is it fully factorised?
- A. Yes
- B. No, it should be 8(2x + 1)
Question 15
Factorise fully: 4xy + 32x
- A. 4y(x + 8)
- B. 4(xy + 8x)
- C. 4x(y + 32)
- D. 4x(y + 8)
Question 16
Factorise β5x β 25 by taking out a negative common factor.
- A. β5(x + 25)
- B. β5(x + 5)
- C. β5(x β 5)
- D. 5(x + 5)
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