Factorising Harder Quadratics - Worksheets, Questions and Revision

20 original exam-style questions - 6 pages of questions with a full mark scheme - free printable PDF.

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7.5 Factorising Harder Quadratics

EDEXCEL 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Factorise fully.

(a)x² + 7x + 10(2)

(b)x² - 3x - 18(2)

(Total for Question 1 is 4 marks)

Factorise, using the difference of two squares.

(a)x² - 49(1)

(b)4x² - 25(2)

(Total for Question 2 is 3 marks)

This question is about the quadratic expression x² + 10x + 25 and the equation x² - 9x + 20 = 0.

(a)Factorise x² + 10x + 25(2)

(b)Solve x² - 9x + 20 = 0(3)

(Total for Question 3 is 5 marks)

Factorise fully.

(a)2x² + 7x + 3(2)

(b)3x² - 5x - 2(2)

(Total for Question 4 is 4 marks)

Factorise fully.

(a)6x² + 13x + 6(3)

(b)4x² - 12x + 9(2)

(Total for Question 5 is 5 marks)

Simplify fully (x² - 4) / (x² + 5x + 6)

(Total for Question 6 is 4 marks)

Solve 2x² - 5x - 3 = 0 by factorising.

(Total for Question 7 is 3 marks)

Factorise fully 3x² + 6x - 45

(Total for Question 8 is 3 marks)

Factorise 9x² - 16

(Total for Question 9 is 2 marks)

Solve x² = 8x - 15

(Total for Question 10 is 3 marks)

Factorise fully 12x² - 17x + 6

(Total for Question 11 is 3 marks)

Simplify fully (2x² + 5x - 3) / (x² - 9)

(Total for Question 12 is 4 marks)

Solve 6x² - 5x - 6 = 0 by factorising.

(Total for Question 13 is 3 marks)

A rectangle has width x cm. Its length is 3 cm more than twice its width. The area of the rectangle is 90 cm².

(a)Show that 2x² + 3x - 90 = 0(2)

(b)Solve the equation to find x, and hence find the dimensions of the rectangle.(4)

(Total for Question 14 is 6 marks)

Factorise 4x² - 9y²

(Total for Question 15 is 2 marks)

Solve (2x - 3)² = 25, using factorisation.

(Total for Question 16 is 3 marks)

Factorise fully 6x² - 24

(Total for Question 17 is 2 marks)

Solve 10x² - 3x - 4 = 0 by factorising.

(Total for Question 18 is 3 marks)

x and y are positive integers such that x² - y² = 45 and x - y = 5.

(Total for Question 19 is 4 marks)

n is a positive integer. Two consecutive odd numbers can be written as (2n + 1) and (2n - 1). Prove that the difference of their squares is always a multiple of 8.

(Total for Question 20 is 3 marks)

Mark scheme · 7.5 Factorising Harder Quadratics

Question 1

(a) M1 identifies a pair of numbers with sum 7 and product 10 (2 and 5), or one correct bracket seen
(a) A1 (x + 2)(x + 5) cao
(a) Answer: (x + 2)(x + 5)
(b) M1 identifies a pair of numbers with sum -3 and product -18 (-6 and 3), or one correct bracket seen
(b) A1 (x - 6)(x + 3) cao
(b) Answer: (x - 6)(x + 3)

Question 2

(a) B1 (x - 7)(x + 7) cao
(a) Answer: (x - 7)(x + 7)
(b) M1 recognises √4x² = 2x and √25 = 5, one correct bracket seen
(b) A1 (2x - 5)(2x + 5) cao
(b) Answer: (2x - 5)(2x + 5)

Question 3

(a) M1 identifies repeated factor 5 (sum 10, product 25)
(a) A1 (x + 5)² oe e.g. (x + 5)(x + 5)
(a) Answer: (x + 5)²
(b) M1 identifies a pair of numbers with sum -9 and product 20 (-4 and -5)
(b) M1 (x - 4)(x - 5) = 0 seen, or ft their factorisation
(b) A1 x = 4 and x = 5 oe, both values required
(b) Answer: x = 4 or x = 5

Question 4

(a) M1 splits middle term correctly, e.g. 2x² + 6x + x + 3, or one correct bracket seen
(a) A1 (2x + 1)(x + 3) cao
(a) Answer: (2x + 1)(x + 3)
(b) M1 splits middle term correctly, e.g. 3x² - 6x + x - 2, or one correct bracket seen
(b) A1 (3x + 1)(x - 2) cao
(b) Answer: (3x + 1)(x - 2)

Question 5

(a) M1 finds two numbers with product 36 (6 x 6) and sum 13 (9 and 4)
(a) M1 splits middle term correctly, e.g. 6x² + 9x + 4x + 6, and groups into two pairs
(a) A1 (2x + 3)(3x + 2) cao
(a) Answer: (2x + 3)(3x + 2)
(b) M1 recognises perfect square, √4x²=2x, √9=3
(b) A1 (2x - 3)² oe
(b) Answer: (2x - 3)²

Question 6

M1 factorises numerator: (x - 2)(x + 2)
M1 factorises denominator: (x + 2)(x + 3)
M1 cancels common factor (x + 2), ft from their factorisations
A1 (x - 2) / (x + 3) cao
Answer: (x - 2) / (x + 3)

Question 7

M1 finds two numbers with product -6 (2 x -3) and sum -5 (-6 and 1)
M1 (2x + 1)(x - 3) = 0 seen, or ft their factorisation
A1 x = -1/2 and x = 3 oe, both values required
Answer: x = -1/2 or x = 3

Question 8

M1 takes out common factor of 3: 3(x² + 2x - 15)
M1 identifies a pair of numbers with sum 2 and product -15 (5 and -3)
A1 3(x + 5)(x - 3) cao, fully factorised
Answer: 3(x + 5)(x - 3)

Question 9

M1 recognises √9x²=3x and √16=4, one correct bracket seen
A1 (3x - 4)(3x + 4) cao
Answer: (3x - 4)(3x + 4)

Question 10

M1 rearranges to x² - 8x + 15 = 0
M1 (x - 3)(x - 5) = 0 seen, or ft their rearrangement
A1 x = 3 and x = 5 oe, both values required
Answer: x = 3 or x = 5

Question 11

M1 finds two numbers with product 72 (12 x 6) and sum -17 (-8 and -9)
M1 splits middle term correctly, e.g. 12x² - 8x - 9x + 6, and groups into two pairs
A1 (3x - 2)(4x - 3) cao
Answer: (3x - 2)(4x - 3)

Question 12

M1 factorises numerator: (2x - 1)(x + 3)
M1 factorises denominator: (x - 3)(x + 3)
M1 cancels common factor (x + 3), ft from their factorisations
A1 (2x - 1) / (x - 3) cao
Answer: (2x - 1) / (x - 3)

Question 13

M1 finds two numbers with product -36 (6 x -6) and sum -5 (-9 and 4)
M1 (3x + 2)(2x - 3) = 0 seen, or ft their factorisation
A1 x = -2/3 and x = 3/2 oe, both values required
Answer: x = -2/3 or x = 3/2

Question 14

(a) M1 sets up area = x(2x + 3) = 90
(a) A1 cso: correctly rearranges to 2x² + 3x - 90 = 0, no errors seen
(a) Answer: 2x² + 3x - 90 = 0 (shown)
(b) M1 finds two numbers with product -180 (2 x -90) and sum 3 (15 and -12)
(b) M1 (2x + 15)(x - 6) = 0 seen, or ft their factorisation
(b) dM1 rejects x = -7.5 as width cannot be negative, dependent on previous M mark
(b) A1 width = 6 cm and length = 15 cm, both values required with units
(b) Answer: x = 6; width = 6 cm, length = 15 cm

Question 15

M1 recognises √4x²=2x and √9y²=3y, one correct bracket seen
A1 (2x - 3y)(2x + 3y) cao
Answer: (2x - 3y)(2x + 3y)

Question 16

M1 rearranges to (2x - 3)² - 25 = 0 and writes as difference of two squares [(2x-3)-5][(2x-3)+5] = 0
M1 simplifies to (2x - 8)(2x + 2) = 0, or ft equivalent
A1 x = 4 and x = -1 oe, both values required
Answer: x = 4 or x = -1

Question 17

M1 takes out common factor of 6: 6(x² - 4)
A1 6(x - 2)(x + 2) cao, fully factorised
Answer: 6(x - 2)(x + 2)

Question 18

M1 finds two numbers with product -40 (10 x -4) and sum -3 (5 and -8)
M1 (5x - 4)(2x + 1) = 0 seen, or ft their factorisation
A1 x = 4/5 and x = -1/2 oe, both values required
Answer: x = 4/5 or x = -1/2

Question 19

M1 factorises using difference of two squares: (x - y)(x + y) = 45
M1 substitutes x - y = 5 to obtain 5(x + y) = 45, so x + y = 9
M1 solves the pair of linear equations x - y = 5 and x + y = 9 simultaneously
A1 x = 7 and y = 2, both values required
Answer: x = 7, y = 2

Question 20

M1 writes the difference of two squares (2n+1)² - (2n-1)² and factorises as [(2n+1)-(2n-1)][(2n+1)+(2n-1)]
M1 simplifies brackets correctly to (2)(4n)
A1 cso: concludes = 8n, which is a multiple of 8 for all integer n, fully justified
Answer: (2n+1)² - (2n-1)² = 8n, a multiple of 8 (proven)

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Factorising Harder Quadratics - Worksheets, Questions and Revision

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7.5 Factorising Harder Quadratics

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Question 9

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Question 11

Question 12

Question 13

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Question 19

Question 20

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