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GCSE · Algebra

5.7D Expanding and Factorising Quadratics: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Expand each expression.

(a)Expand 4(x + 6)(1)

(b)Expand 5(2x - 3)(1)

(Total for Question 1 is 2 marks)

Expand and simplify each expression.

(a)Expand and simplify 3(x + 4) + 2(x + 1)(2)

(b)Expand and simplify 6(x - 2) - 3(x - 5)(2)

(Total for Question 2 is 4 marks)

Expand and simplify each expression.

(a)(x + 3)(x + 8)(2)

(b)(x + 5)(x + 5)(2)

(Total for Question 3 is 4 marks)

Which of these is the correct expansion of (x - 4)(x + 9)?

x² + 5x - 36
x² - 5x - 36
x² + 5x + 36
x² - 13x - 36

(Total for Question 4 is 1 mark)

Factorise each expression.

(a)Factorise x² + 7x + 12(2)

(b)Factorise x² + 3x - 18(2)

(Total for Question 5 is 4 marks)

Expand and simplify each expression.

(a)(x - 4)(x + 9)(2)

(b)(2x + 1)(x - 5)(2)

(c)(3x - 2)(2x - 5)(3)

(Total for Question 6 is 7 marks)

Expand and simplify each expression.

(a)(x + 9)(x - 9)(2)

(b)(x - 7)(x - 7)(2)

(Total for Question 7 is 4 marks)

Factorise each expression.

(a)Factorise x² - 5x - 24(2)

(b)Factorise x² - 13x + 40(2)

(Total for Question 8 is 4 marks)

Factorise each expression.

(a)Factorise x² - 49(2)

(b)Factorise x² - 144(2)

(Total for Question 9 is 4 marks)

Factorise fully each expression.

(a)Factorise fully 2x² + 14x + 20(3)

(b)Factorise fully 5x² - 45(3)

(Total for Question 10 is 6 marks)

Show that (x + 4)(x - 6) + 5 = x² - 2x - 19

(Total for Question 11 is 2 marks)

Solve x² + 2x - 24 = 0

(Total for Question 12 is 3 marks)

Solve x² + 7x = 30

(Total for Question 13 is 4 marks)

A rectangular lawn has length (x + 11) m and width (x - 2) m. The area of the lawn is 68 m^2.

(a)Show that x² + 9x - 90 = 0(3)

(b)Solve the equation to find the value of x, given that x > 0(3)

(c)Hence work out the perimeter of the lawn.(2)

(Total for Question 14 is 8 marks)

Factorise each expression.

(a)Factorise 3x² + 13x + 4(3)

(b)Factorise 4x² + 4x - 3(3)

(Total for Question 15 is 6 marks)

Prove algebraically that (2n + 3)² - (2n - 3)² is always a multiple of 24, for any positive integer n.

(Total for Question 16 is 4 marks)

Simplify fully (x² - 16) / (x² + x - 12)

(Total for Question 17 is 3 marks)

Three consecutive positive integers are such that the product of the smallest and the largest is 23 more than 5 times the middle integer. Find the three integers.

(Total for Question 18 is 6 marks)

Solve algebraically: (x + 4)(x - 3) = 2(x + 4)
You must show your working.

(Total for Question 19 is 5 marks)

The expression x² + kx + 36 can be factorised as (x + a)(x + b), where a and b are integers and k is negative. List all possible values of k.

(Total for Question 20 is 4 marks)

Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.

(Total for Question 21 is 4 marks)

Mark scheme · 5.7D Expanding and Factorising Quadratics: Fluency and Exam Drill

Question 1

(a) B1 4x + 24 oe
(a) Answer: 4x + 24
(b) B1 10x - 15 oe
(b) Answer: 10x - 15

Question 2

(a) M1 correct expansion of at least one bracket, 3x + 12 or 2x + 2
(a) A1 5x + 14 cao
(a) Answer: 5x + 14
(b) M1 correct expansion of at least one bracket, 6x - 12 or -3x + 15
(b) A1 3x + 3 cao
(b) Answer: 3x + 3

Question 3

(a) M1 attempts expansion, e.g. x² + 8x + 3x + 24
(a) A1 x² + 11x + 24 cao
(a) Answer: x² + 11x + 24
(b) M1 attempts expansion, e.g. x² + 5x + 5x + 25
(b) A1 x² + 10x + 25 cao
(b) Answer: x² + 10x + 25

Question 4

B1 x² + 5x - 36
Answer: x² + 5x - 36

Question 5

(a) M1 attempts a factor pair of 12 that sums to 7, e.g. 3 and 4
(a) A1 (x + 3)(x + 4) cao
(a) Answer: (x + 3)(x + 4)
(b) M1 attempts a factor pair of -18 that sums to 3, e.g. 6 and -3
(b) A1 (x + 6)(x - 3) cao
(b) Answer: (x + 6)(x - 3)

Question 6

(a) M1 attempts expansion, e.g. x² + 9x - 4x - 36
(a) A1 x² + 5x - 36 cao
(a) Answer: x² + 5x - 36
(b) M1 attempts expansion, e.g. 2x² - 10x + x - 5
(b) A1 2x² - 9x - 5 cao
(b) Answer: 2x² - 9x - 5
(c) M1 attempts expansion to 4 terms, e.g. 6x² - 15x - 4x + 10
(c) M1 collects the two middle terms correctly (ft), -15x - 4x = -19x
(c) A1 6x² - 19x + 10 cao
(c) Answer: 6x² - 19x + 10

Question 7

(a) M1 attempts expansion, e.g. x² - 9x + 9x - 81
(a) A1 x² - 81 cao
(a) Answer: x² - 81
(b) M1 attempts expansion, e.g. x² - 7x - 7x + 49
(b) A1 x² - 14x + 49 cao
(b) Answer: x² - 14x + 49

Question 8

(a) M1 attempts a factor pair of -24 that sums to -5, e.g. -8 and 3
(a) A1 (x - 8)(x + 3) cao
(a) Answer: (x - 8)(x + 3)
(b) M1 attempts a factor pair of 40 that sums to -13, e.g. -8 and -5
(b) A1 (x - 8)(x - 5) cao
(b) Answer: (x - 8)(x - 5)

Question 9

(a) M1 recognises the difference of two squares
(a) A1 (x - 7)(x + 7) cao
(a) Answer: (x - 7)(x + 7)
(b) M1 recognises the difference of two squares
(b) A1 (x - 12)(x + 12) cao
(b) Answer: (x - 12)(x + 12)

Question 10

(a) M1 takes out the common factor 2, 2(x² + 7x + 10)
(a) M1 factorises the quadratic, (x + 2)(x + 5)
(a) A1 2(x + 2)(x + 5) cao, fully factorised
(a) Answer: 2(x + 2)(x + 5)
(b) M1 takes out the common factor 5, 5(x² - 9)
(b) M1 recognises the difference of two squares
(b) A1 5(x - 3)(x + 3) cao, fully factorised
(b) Answer: 5(x - 3)(x + 3)

Question 11

M1 expands (x + 4)(x - 6) correctly, e.g. x² - 6x + 4x - 24 = x² - 2x - 24
A1 adds 5 to reach x² - 2x - 19, matching the given expression (cso)
Answer: x² - 2x - 19 (shown)

Question 12

M1 factorises to (x + 6)(x - 4) = 0 (or uses a correct alternative method)
A1 x = -6
A1 x = 4 cao
Answer: x = -6 or x = 4

Question 13

M1 rearranges to x² + 7x - 30 = 0
M1 factorises to (x + 10)(x - 3) = 0
A1 x = -10
A1 x = 3 cao
Answer: x = -10 or x = 3

Question 14

(a) M1 expands (x + 11)(x - 2) correctly, e.g. x² - 2x + 11x - 22 = x² + 9x - 22
(a) M1 sets the expansion equal to 68 and rearranges
(a) A1 correctly shows x² + 9x - 90 = 0 (cso)
(a) Answer: x² + 9x - 90 = 0 (shown)
(b) M1 factorises to (x + 15)(x - 6) = 0 (or uses a correct alternative method)
(b) A1 x = -15 and x = 6
(b) A1 selects x = 6, since x > 0 cao
(b) Answer: x = 6
(c) M1 finds length = 17 m and width = 4 m (ft their x)
(c) A1 42 m cao
(c) Answer: 42 m

Question 15

(a) M1 identifies factors of 3 to try for the x-coefficients, e.g. 3 and 1
(a) M1 identifies a factor pair of 4 that gives the correct middle term on trial, e.g. 1 and 4
(a) A1 (3x + 1)(x + 4) cao
(a) Answer: (3x + 1)(x + 4)
(b) M1 identifies factors of 4 to try for the x-coefficients, e.g. 2 and 2
(b) M1 identifies a factor pair of -3 that gives the correct middle term on trial, e.g. 3 and -1
(b) A1 (2x + 3)(2x - 1) cao
(b) Answer: (2x + 3)(2x - 1)

Question 16

M1 expands (2n + 3)² correctly to 4n² + 12n + 9
M1 expands (2n - 3)² correctly to 4n² - 12n + 9
A1 subtracts to obtain 24n (fully simplified)
B1 correct conclusion, e.g. since 24n = 24 x n, the expression is always a multiple of 24 for integer n
Answer: (2n+3)² - (2n-3)² = 24n, which is always a multiple of 24

Question 17

M1 factorises the numerator, (x - 4)(x + 4)
M1 factorises the denominator, (x + 4)(x - 3)
A1 (x - 4) / (x - 3) cao, fully cancelled
Answer: (x - 4) / (x - 3)

Question 18

M1 lets the middle integer be n and writes the smallest as n - 1 and the largest as n + 1 (or equivalent)
M1 forms the equation (n - 1)(n + 1) = 5n + 23
M1 expands and rearranges to the 3-term quadratic n² - 5n - 24 = 0
M1 factorises to (n - 8)(n + 3) = 0
A1 n = 8 (rejects n = -3 as not a positive integer)
A1 states the three integers 7, 8 and 9 cao
Answer: 7, 8 and 9

Question 19

M1 expands (x + 4)(x - 3) correctly to x² + x - 12
M1 rearranges to form x² - x - 20 = 0 (does not simply cancel the common factor (x + 4) from both sides)
M1 factorises correctly to (x - 5)(x + 4) = 0
A1 x = 5
A1 x = -4 (both solutions identified) cao
Answer: x = 5 or x = -4

Question 20

M1 lists factor pairs of 36, e.g. 1x36, 2x18, 3x12, 4x9, 6x6
M1 recognises that since the constant term is positive and k is negative, a and b must both be negative, so k = -(a + b)
A1 at least three correct values from -12, -13, -15, -20, -37
A1 all five correct values stated, k = -12, -13, -15, -20, -37 cao
Answer: k = -12, -13, -15, -20 or -37

Question 21

B1 defines two consecutive odd numbers algebraically, e.g. 2n + 1 and 2n + 3
M1 applies the difference of two squares, (2n+3)² - (2n+1)² = [(2n+3)+(2n+1)][(2n+3)-(2n+1)]
A1 simplifies correctly to 8n + 8, or 8(n + 1)
B1 correct conclusion, e.g. since 8(n+1) is 8 multiplied by an integer, the difference is always a multiple of 8
Answer: (2n+3)² - (2n+1)² = 8(n+1), which is always a multiple of 8

Expanding and Factorising Quadratics: Fluency and Exam Drill - Worksheets, Questions and Revision

5.7D Expanding and Factorising Quadratics: Fluency and Exam Drill

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21