A rectangle has area (2x2 + x - 6) cm2. One side of the rectangle has length (x + 2) cm.
(a)Show that the length of the other side is (2x - 3) cm.(2)
(b)Given that x = 6, work out the perimeter of the rectangle.(2)
(Total for Question 17 is 4 marks)
18
A student is asked to factorise 4x2 - 16 fully. The student writes: 4x2 - 16 = (2x - 4)(2x + 4).
(a)Explain why the student's answer is not fully factorised, even though it is mathematically correct.(1)
(b)Write down the fully factorised form of 4x2 - 16.(1)
(Total for Question 18 is 2 marks)
19
Two positive integers, p and q, with p greater than q, satisfy p2 - q2 = 45. Given also that p - q = 3, find the value of p and the value of q.
(Total for Question 19 is 4 marks)
20
Solve 3x2 = 13x - 4 by factorising. Show your working.
(Total for Question 20 is 4 marks)
21
Simplify fully (2x2 + 7x + 3) / (x2 - 9)
(Total for Question 21 is 3 marks)
22
A rectangle has width w cm and length (w + 3) cm. The area of the rectangle is 40 cm2 greater than 6 times its width. Form and solve a quadratic equation to find the width of the rectangle.
(Total for Question 22 is 4 marks)
23
Factorise fully 6x2 + xy - 12y2, giving your answer in terms of x and y.
(Total for Question 23 is 3 marks)
24
By substituting y = x2 - 2x, or otherwise, solve (x2 - 2x)2 - 5(x2 - 2x) - 24 = 0, giving all real solutions for x.
(Total for Question 24 is 4 marks)
Mark scheme · 7.5D Factorising Harder Quadratics: Fluency and Exam Drill
Question 1
B1 (x - 7)(x + 7) cao
Answer: (x - 7)(x + 7)
Question 2
M1 identifies a pair of numbers with product 6 and sum 5 (2 and 3), or one correct bracket seen
A1 (2x + 3)(x + 1) cao
Answer: (2x + 3)(x + 1)
Question 3
M1 identifies a pair of numbers with product 6 and sum -7 (-6 and -1), or one correct bracket seen
A1 (2x - 1)(x - 3) cao
Answer: (2x - 1)(x - 3)
Question 4
M1 identifies a pair of numbers with product 9 and sum 10 (9 and 1), or one correct bracket seen
A1 (3x + 1)(x + 3) cao
Answer: (3x + 1)(x + 3)
Question 5
M1 identifies a pair of numbers with product -30 and sum -13 (-15 and 2), or one correct bracket seen
A1 (5x + 2)(x - 3) cao
Answer: (5x + 2)(x - 3)
Question 6
M1 recognises √4x2 = 2x and √9 = 3, one correct bracket seen
A1 (2x - 3)(2x + 3) cao
Answer: (2x - 3)(2x + 3)
Question 7
M1 recognises √9x2 = 3x and √25 = 5, one correct bracket seen
A1 (3x - 5)(3x + 5) cao
Answer: (3x - 5)(3x + 5)
Question 8
M1 identifies a pair of numbers with product -18 and sum 7 (9 and -2), or one correct bracket seen
A1 (3x - 1)(2x + 3) cao
Answer: (3x - 1)(2x + 3)
Question 9
M1 identifies a pair of numbers with product -30 and sum -1 (5 and -6), or one correct bracket seen
A1 (2x + 5)(x - 3) cao
Answer: (2x + 5)(x - 3)
Question 10
M1 identifies a pair of numbers with product -36 and sum -5 (4 and -9), or one correct bracket seen
A1 (3x + 4)(x - 3) cao
Answer: (3x + 4)(x - 3)
Question 11
M1 identifies a pair of numbers with product -12 and sum 4 (6 and -2), or one correct bracket seen
A1 (2x + 3)(2x - 1) cao
Answer: (2x + 3)(2x - 1)
Question 12
M1 identifies a pair of numbers with product -24 and sum 2 (6 and -4), or one correct bracket seen
A1 (4x + 3)(2x - 1) cao
Answer: (4x + 3)(2x - 1)
Question 13
M1 identifies a pair of numbers with product -40 and sum -3 (5 and -8), or one correct bracket seen
A1 (5x - 4)(2x + 1) cao
Answer: (5x - 4)(2x + 1)
Question 14
M1 takes out common factor of 3: 3(2x2 + 5x + 2)
M1 identifies a pair of numbers with product 4 and sum 5 (4 and 1), or one correct bracket seen
A1 3(2x + 1)(x + 2) cao, fully factorised
Answer: 3(2x + 1)(x + 2)
Question 15
M1 identifies a pair of numbers with product 6 and sum 7 (6 and 1), leading to (2x + 1)(x + 3)
M1 (2x + 1)(x + 3) = 0 seen, or ft their factorisation, with an attempt to solve
A1 x = -1/2 and x = -3 oe, both values required
Answer: x = -1/2 or x = -3
Question 16
M1 identifies a pair of numbers with product -12 and sum -11 (-12 and 1), leading to (3x + 1)(x - 4)
M1 (3x + 1)(x - 4) = 0 seen, or ft their factorisation, with an attempt to solve
A1 x = -1/3 and x = 4 oe, both values required
Answer: x = -1/3 or x = 4
Question 17
(a) M1 factorises 2x2 + x - 6 as (x + 2)(2x - 3), or divides 2x2 + x - 6 by (x + 2)
(a) A1 cso: correctly shows the other side is (2x - 3) cm, no errors seen
(a) Answer: 2x - 3 cm (shown)
(b) M1 substitutes x = 6 into both (x + 2) and (2x - 3) to find 8 cm and 9 cm, or uses P = 2(x + 2) + 2(2x - 3) with x = 6
(b) A1 34 cm cao
(b) Answer: 34 cm
Question 18
(a) B1 identifies that each bracket, (2x - 4) and (2x + 4), still has a common factor of 2 that has not been taken out, oe
(a) Answer: Each bracket has a common factor of 2, so the expression has not been factorised fully.
(b) B1 4(x - 2)(x + 2) oe cao
(b) Answer: 4(x - 2)(x + 2)
Question 19
M1 uses the difference of two squares to write p2 - q2 = (p - q)(p + q)
M1 substitutes p - q = 3 into 3(p + q) = 45 to obtain p + q = 15
M1 solves the simultaneous equations p - q = 3 and p + q = 15
A1 p = 9 and q = 6, both values required
Answer: p = 9, q = 6
Question 20
M1 rearranges to 3x2 - 13x + 4 = 0 oe
M1 identifies a pair of numbers with product 12 and sum -13 (-12 and -1), leading to (3x - 1)(x - 4)
A1 x = 4 cao
A1 x = 1/3 oe, ft from a correctly factorised bracket
Answer: x = 4 or x = 1/3
Question 21
M1 factorises numerator: (2x + 1)(x + 3)
M1 factorises denominator: (x - 3)(x + 3)
A1 (2x + 1) / (x - 3) cao, after cancelling the common factor (x + 3)
Answer: (2x + 1) / (x - 3)
Question 22
M1 forms the equation w(w + 3) = 6w + 40 oe
M1 rearranges to w2 - 3w - 40 = 0 oe
M1 factorises and solves to obtain w = 8 or w = -5
A1 w = 8 cm cao, with the negative solution rejected since width cannot be negative
Answer: w = 8 cm
Question 23
M1 attempts to factorise into two linear brackets in x and y, e.g. finds a factor pair for 6 and a factor pair for -12 that combine to give a coefficient of 1 for xy
M1 writes down brackets with correct x and y terms, allowing at most one sign error, e.g. (2x + 3y)(3x - 4y)
A1 (2x + 3y)(3x - 4y) cao
Answer: (2x + 3y)(3x - 4y)
Question 24
M1 substitutes y = x2 - 2x to rewrite the equation as y2 - 5y - 24 = 0
M1 factorises and solves to obtain y = 8 or y = -3
dM1 forms x2 - 2x - 8 = 0 from y = 8 and factorises/solves it, dependent on both previous M marks
A1 x = 4 and x = -2 only cao, with the y = -3 branch correctly rejected since x2 - 2x + 3 = 0 has discriminant 4 - 12 = -8, which is negative, so gives no real solutions