Algebraic manipulation — Paper 2 · IGCSE Mathematics 0580

18 5 − 2 4 2 C = D = 1 4 − 1 5 (a) Write as a single matrix (i) C − 3 D, Answer(a)(i) [2] (ii) CD. Answer(a)(ii) [2] (b) Find C −.1 Answer(b) [2]

6 marks

Mark scheme: 18  − 7 − 8  2 B1 any 2 correct (a)    4 − 11  22 0  (b) 2 B1 either column correct    0 22  1  4 2  2 M1 either adjoint matrix correct or (c) o.e   determinant 22 seen 22   − 1 5

This question in 0580/21 Oct/Nov 2004

 2 3  5 A =  _   4 5  Find A–1, the inverse of the matrix A.     Answer       [2]

2 marks

Mark scheme: 5 1  5 − 3   5.2 − 5.1  2 M1 det A or |A| or 5×–2 – 4×–3 = 2 or  or     1  5 − 3  a b  2  4 − 2  2 − 1      or  seen  4 − 2  2  c d  Allow 5/2, –3/2, 4/2, –2/2 in matrix 7

This question in 0580/21 May/June 2009

6 0 1 7 1 A = B = − 8 − 4 0 − 5 Calculate the value of 5 |A| + |B|, where |A| and |B| are the determinants of A and B. Answer [2]

2 marks

Mark scheme: 6 5 2 M1 |A| = 0 × –4 – 1 × –8 or better or |B| = 7 × –5 – 0 × 1 or better det symbol can be implied by the working

This question in 0580/21 Oct/Nov 2009

12 Simplify 16 – 4(3x – 2)2. Answer [3]

3 marks

Mark scheme: 12 –36x2 + 48x or 12x(4 – 3x) oe 3 M1 squaring to “9x2 –12x + 4” algebraic or other partly factorised versions M1 multiplying by –4 terms M1 adding 16 only

This question in 0580/21 Oct/Nov 2009

5d + 4w Examiner's 11 Make d the subject of the formula c = . Use 2w Answer d = [3]

3 marks

Mark scheme: 2cw 4 w 11 oe 3 M1 one correct move to clear fractions 5 M1 second correct move to subtract term M1 third correct move dividing by 5 May be in any order

This question in 0580/21 May/June 2010

16 Simplify Examiner's Use 0.75  p 4  (a)   ,  16  Answer(a) [2] (b) 32q-3 ÷ 23q-2. Answer(b) [2]

4 marks

Mark scheme: p 16 (a) or 0.125p3 1, 1 Independent marks for letter and no. 8 9 (b) q–1 1, 1 Independent marks for letter and no. 8 1 9 Allow 1 q–1 or 8 8q

This question in 0580/21 May/June 2010

21 (a) A is a (2 × 4) matrix, B is a (3 × 2) matrix and C is a (1 × 3) matrix. Examiner's Use Which two of the following matrix products is it possible to work out? A2 B2 C2 AB AC BA BC CA CB Answer(a) and [2]  1 3    2 4 (b) Find the inverse of   .  1 1     8 4  Simplify your answer as far as possible.     Answer(b)       [3]  4 2  (c) Explain why the matrix   does not have an inverse.  6 3    Answer(c) [1]

6 marks

Mark scheme: 21 (a) CB and BA cao 1, 1 Independent  8 − 24  1 1 3 1 1 (b)  − 4 16  cao 3 M1 2 × 4 − 4 × 8 (= 32 )=  1 3   −  M1  4 4  seen  − 1 1   8 2  (c) determinant is zero 1 Allow cannot divide by zero

This question in 0580/21 May/June 2010

10 Make x the subject of the formula. Examiner's Use x + 3 P = x Answer x = [4]

4 marks

Mark scheme: 3 10 x = 4 M1 for each of the four moves completed P − 1 correctly

This question in 0580/22 May/June 2010

13 Examiner's  6 − 3  x  Use M =    .  4 5  1  (a) Find the matrix M. Answer(a) M = [2] (b) Simplify ( x 1 ) M. Answer(b) [2]

4 marks

Mark scheme: 13 (a)  6 x − 3  but not  6 x − 3  2 B1 6x – 3 or B1 4x + 5 in a (2 × 1) matrix on  4 x + 5   4 x ( + )5  answer line (b) (6x2 + x + 5) cao 2 M1 any 1 × 1 matrix in answer space

This question in 0580/22 May/June 2010

12 Expand and simplify 2(x – 3)2 – (2x – 3)2. Examiner's Use Answer [3]

3 marks

Mark scheme: 12 9 – 2x2 3 B1 for x2 – 3x – 3x + 9 or 2x2 – 6x – 6x + 18 B1 for 4x2 – 6x – 6x + 9 or –4x2 + 6x + 6x – 9

This question in 0580/23 May/June 2010

r ( y + 2 ) Examiner's 16 Make y the subject of the formula. A = Use 5 Answer y = [3]

3 marks

Mark scheme: 5 A 5 A 2 r 16 − 2 or 3 M1 for correctly multiplying by 5 r r M1 for correctly dividing by r M1 for correct subtraction in any order 40 2

This question in 0580/23 May/June 2010

3 −1 Examiner's23 A = (1 4 ) B = Use ( −2 2 ) Find (a) AB, Answer(a) AB = [2] (b) the inverse matrix B–1 , Answer(b) B–1 = [2] (c) BB–1. Answer(c) BB–1 = [1]

5 marks

Mark scheme: 23 (a) (− 5 7 ) 2 B1 either correct in a (1 × 2) matrix 1  2 1   2 1  (b) oe 2 M1 for seen or 2 × 3 – –1 × –2 ( = 4) 2 3  4  2 3    1 0  (c)  0 1  or I cao 1

This question in 0580/23 May/June 2010

16 Simplify this fraction. For Examiner's 2 Use x − 5 x + 6 x 2 − 4 Answer [4]

4 marks

Mark scheme: 16 x − 3 4 B2 (x – 3)(x – 2) or B1 (x + a)(x + b) x + 2 where ab = 6 or a + b = –5 B1 (x – 2)(x + 2)  8 0 

This question in 0580/21 Oct/Nov 2010

17  2 2  A =   2 −2   Work out (a) A2,       Answer(a)   [2]     (b) A–1, the inverse of A.       Answer(b)   [2]    

4 marks

Mark scheme: 17 (a)  8 0  oe 2 B1 for one column (or row) correct  0 8  1 1   a c  4 4   (b) oe 2 B1 for –1/8  seen 1 1 d    b  or B1 for  −− 22 − 22   4 −4 

This question in 0580/21 Oct/Nov 2010

3 Rearrange the formula J = mv – mu to make m the subject. Answer m = [2]

2 marks

Mark scheme: J 3 m = 2 M1 m(v – u) seen v − u

This question in 0580/22 Oct/Nov 2010

18 For  2 4   3 −4  Examiner's A =   B =   Use      5 3   −5 2  (a) Work out AB. Answer(a) [2] (b) Find | B |, the determinant of B. Answer(b) [1] (c) I is the (2 × 2) identity matrix. Find the matrix C, where C = A – 7I . Answer(c) [2]

5 marks

Mark scheme:  14 0  18 (a)  0 −14  2 B1 two or three correct answers (b) –14 1  − 5 4  (c)  5 − 4  2 B1 two or three terms correct 2 2 2

This question in 0580/22 Oct/Nov 2010

1 2 For 24 (a) Write − as a single fraction in its lowest terms. Examiner's y x Use Answer(a) [2] x 2 + x (b) Write in its lowest terms. 3x + 3 Answer(b) [3]

5 marks

Mark scheme: x 2 y 24 (a) 2 B1 correct numerator xy B1 correct denominator x (b) www 3 M1 x(x + 1) M1 3(x + 1) 3 1 2

This question in 0580/23 Oct/Nov 2010

16 ForFor Examiner'sExaminer's UseUse k cm The diagram shows a square of side k cm. The circle inside the square touches all four sides of the square. (a) The shaded area is A cm2. Show that 4A = 4k2 – πk2. Answer (a) [2] (b) Make k the subject of the formula 4A = 4k2 – πk2. Answer(b) k = [3]

5 marks

Mark scheme: k 16 (a) Answer given 2 M1 (A =)k2 – π    2  πk 2 E1 A = k2 – 4 correctly completed to 4A = 4k2 – πk2 4 A A (b) k = (±) or 2 3 M1 factorising (must contain a π) (4 − π ) (4 − π ) M1 division (by coefficient of k2) M1 square root

This question in 0580/21 May/June 2011

1 Factorise completely. For 2xy – 4yz Examiner's Use Answer [2]

2 marks

Mark scheme: Qu. Answers Mark Part Mark 1 2y(x – 2z) 2 B1 for y(2x – 4z) or 2(xy – 2yz)

This question in 0580/23 May/June 2011

x 2 Make x the subject of the formula. y = + 5 3 Answer x = [2]

2 marks

Mark scheme: 2 (x =) 3(y – 5) oe final answer 2 M1 for correct first move x y – 5 = or 3y = x + 15 3 M1 for their correct second move

This question in 0580/23 May/June 2011

11 Work out. 2  2 1  (a)    4 3      Answer(a)   [2]   −1  2 1  (b)    4 3      Answer(b)   [2]  

4 marks

Mark scheme:   11 (a)   2 B1 two or three entries correct  20 13   1 12 − 12  1  a c   3 − 1 oe 2 B1 (b)     B1(k ) b d  2  − 2 1      − 4 2

This question in 0580/21 Oct/Nov 2011

2 Factorise completely ax + bx + ay + by. Answer [2]

2 marks

Mark scheme: 2 (a + b)(x + y) 2 M1 x(a + b) + y(a + b) or M1 a(x + y) + b(x + y)

This question in 0580/22 Oct/Nov 2011

1 18 w = LC (a) Find w when L = 8 × 10 O3 and C = 2 × 10 O9. Give your answer in standard form. Answer(a) w = [3] (b) Rearrange the formula to make C the subject. Answer(b) C = [3] Question 19 is printed on the next page.

6 marks

Mark scheme: 18 (a) 2.5 × 105 3 B2 250000 oe or M1 correct part value seen (b) C = 1/(Lw2) 3 M1 each correct move

This question in 0580/22 Oct/Nov 2011

11 Factorise completely. For p2x – 4q2x Examiner's Use Answer [3]

3 marks

Mark scheme: 11 x(p – 2q)(p + 2q) 3 M2 for (px – 2qx)(p + 2q) or (p – 2q)(px + 2qx) or M1 for x(p2 – 4q2)

This question in 0580/23 Oct/Nov 2011

15 ap = px + c Write p in terms of a, c and x. Answer p = [3]

3 marks

Mark scheme: 15 c 3 M1 one correct move p = M1 second correct move a − x M1 third correct move marked on answer line IGCSE – October/November 2011 0580 23

This question in 0580/23 Oct/Nov 2011

 2  For 20 (a) N =   . The order of the matrix N is 2 × 1. Examiner's  6  Use P = (1 3). The order of the matrix P is 1 × 2. (i) Write down the order of the matrix NP. Answer(a)(i) [1] (ii) Calculate PN. Answer(a)(ii) [1]  2 3  (b) M =   .  2 4  Find M–1, the inverse of M . Answer(b) M–1 = [2]

4 marks

Mark scheme: 20 (a) (i) 2 × 2 1 (ii) (20) 1 Brackets essential 1  4 − 3  1  a b   4 − 3  (b) seen oe 2 M1 for     or k  c d  2 2     − 2 2   − 2 2 2 7 2 + 4 5 2 − 5 2

This question in 0580/23 Oct/Nov 2011

3 Factorise completely. 15p2 + 24pt Answer [2]

2 marks

Mark scheme: 3 3p(5p + 8t) final answer 2 B1 for answer of 3(5p2 + 8pt) or p(15p + 24t) or SC1 for correct answer seen in working 8

This question in 0580/21 May/June 2012

 5 2   − 1 − 2  For Examiner's16 M =   N =    − 3 4   2 6  Use Calculate (a) MN, Answer(a) MN = [2] (b) M−1, the inverse of M. Answer(b) M–1 = [2]

4 marks

Mark scheme:  1 2  16 (a)   2 B1 any two entries correct  11 30  1  4 − 2  1  a b   4 − 2  (b)   oe 2 B1   or k   26  3 5  26  c d   3 5  4 3c

This question in 0580/21 May/June 2012

17 Make w the subject of the formula. 4 + w c = w + 3 Answer w = [4]

4 marks

Mark scheme: 4 − 3c 17 w = www 4 M1 clearing denominator and removing brackets c − 1 M1 correctly collecting terms in w on one side only M1 factorising correctly M1 divide by coefficient of w

This question in 0580/21 May/June 2012

9 Make w the subject of the formula. 3 w t = 2 – a Answer w = [3]

3 marks

Mark scheme: 9 a (2 − t ) 3 M1 correct re-arrangement to isolate the term in w cao oe M1 correct multiplication by a 3 M1 correct division by their 3 An incorrect answer scores a maximum of M2

This question in 0580/23 May/June 2012

13 (a) Find the value of 7p – 3q when p = 8 and q = O5 . Answer(a) [2] (b) Factorise completely. 3uv + 9vw Answer(b) [2]

4 marks

Mark scheme: 13 (a) 71 2 M1 for 7×8 – 3×–5 or B1 56 and –15 (b) 3v (u + 3w) final answer 2 B1 for 3(uv + 3vw) or v (3u + 9w) As final answer

This question in 0580/23 May/June 2012

5 Simplify the expression. 1 1 1 1 2 O b 2 )(a 2 + b 2 ) (a Answer [2]

2 marks

Mark scheme: 5 a(1) – b(1) www cao 2 M1 for a½ a½ – a½ b½ + a½b½ – b½b½ oe

This question in 0580/21 Oct/Nov 2012

19 For Examiner's Use  5 − 4    M =    2 3    Find (a) M2 ,     Answer(a)   [2]     (b) 2M ,     Answer(b)   [1]     (c) |M| , the determinant of M, Answer(c) [1] (d) MO1.     Answer(d)   [2]    

6 marks

Mark scheme: 19 (a) 17 –32 2 M1 any 2 entries correct 16 1 (b) 10 -8 1 4 6 (c) 23 cao 1 (d) 1 3 4 2 M1 3 4 or 1 a b seen 23 –2 5 –2 5 (c) c d IGCSE – October/November 2012 0580 21

This question in 0580/21 Oct/Nov 2012

4 Expand the brackets. y(3 O y3) Answer [2]

2 marks

Mark scheme: 4 3y – y4 final answer 2 B1 for 3y or – y4 as part of two term expression

This question in 0580/23 Oct/Nov 2012

21 Simplify the following. For h 2 − h − 20 Examiner's Use h 2 − 25 Answer [4]  3 2 

4 marks

Mark scheme: 21 h + 4 4 B2 for (h – 5)(h + 4) seen h + 5 B1 for (h – 5)(h + 5) If B2 not scored then SC1 for (h + a)(h + b) where a + b = –1 or ab = –20 1  1 −2  1  a b  1 −2 

This question in 0580/23 Oct/Nov 2012

22 (a) M =   − 1 1  Find M–1, the inverse of M.     Answer(a)   [2]     (b) D, E and X are 2 × 2 matrices. I is the identity 2 × 2 matrix. (i) Simplify DI. Answer(b)(i) [1] (ii) DX = E Write X in terms of D and E. Answer(b)(ii) X = [1]

4 marks

Mark scheme: 1  1 2  1  a b  1 2  22 (a)   2 B1 for   or k   seen 5  1 3  5  c d  1 3  (b)(i) D cao 1 (ii) D–1E cao 1

This question in 0580/23 Oct/Nov 2012

23 f(x) = 3x + 5 g(x) = 4x O 1 For Examiner's Use (a) Find the value of gg(3). Answer(a) [2] (b) Find fg(x), giving your answer in its simplest form. Answer(b)fg(x) = [2] (c) Solve the equation. f –1(x) = 11 Answer(c) x = [1] Question 24 is printed on the next page.

5 marks

Mark scheme: 23 (a) 43 2 M1 for g(11) or 4[4(3) – 1] –1 (b) 12x + 2 2 M1 for 3(4x – 1) + 5 (c) 38 1 2 2 2

This question in 0580/23 Oct/Nov 2012

6 Factorise completely. 12xy – 3x2 Answer … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 6 3x(4y – x) final answer 2 B1 for 3(4xy – x2) or x(12y – 3x)

This question in 0580/21 May/June 2013

10 Factorise completely. ap + bp – 2a – 2b Answer … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 10 (a + b)(p – 2) 2 B1 p(a + b) – 2(a + b) or a(p – 2) + b(p – 2) 4 4 k

This question in 0580/21 May/June 2013

18 (a) Factorise x2 + x – 30. Answer(a) … [2] (x - 5)(x + 4) . (b) Simplify x2 + x - 30 Answer(b) … [1] _____________________________________________________________________________________

3 marks

Mark scheme: 18 (a) (x + 6)(x–5) 2 SC1 for (x + a)(x + b) where ab = –30 or a + b (b) x + 4 1 final answer x + 6 6 k

This question in 0580/21 May/June 2013

22 Write as a single fraction in its simplest form. 2 3 + x + 3 x + 2 Answer … [3] _____________________________________________________________________________________

3 marks

Mark scheme: 22 5 x + 13 3 B1 for common denominator (x + 3)(x + 2) seen oe final answer ( x + 3)( x + 2) M1 for 2(x + 2) + 3(x + 3) soi

This question in 0580/21 May/June 2013

14 (a) Solve 3n + 23 < n + 41. Answer(a) … [2] (b) Factorise completely ab + bc + ad + cd. Answer(b) … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 14 (a) n < 9 2 M1 for 2n < 18 or 2n – 18 < 0 oe If 0 scored SC1 for 9 with incorrect inequality. (b) (b + d)(a + c) 2 B1 for b(a + c) + d(a + c) or a(b + d) + c (b + d)

This question in 0580/23 May/June 2013

20 (a) For Examiner′s 4 Use y = 8 + x Find y when x = 2. Give your answer correct to 4 decimal places. Answer(a) y = … [2] 4 (b) Rearrange y = 8 + to make x the subject. x Answer(b) x = … [4]

6 marks

Mark scheme: 20 (a) [ ± ] 3.1623 cao 2 M1 for √10 seen 4 (b) 2 oe final answer 4 M1 first move completed correctly − 8 y M1 second move completed correctly M1 third move completed correctly M1 final move completed correctly on answer line

This question in 0580/23 May/June 2013

17 For Examiner′s 2 1 5 0 Use M = N = 4 6 1 5 f p f p (a) Work out MN. Answer(a) MN = [2] (b) Find M–1. Answer(b) M–1 = [2] _____________________________________________________________________________________

4 marks

Mark scheme:   17 (a)   2 SC1 for one correct row or column 26 30   1  6 − 1   6 − 1  (b)   oe 2 B1 for k   8  − 4 2   − 4 2  1  a b  or B1 for   8 c d  

This question in 0580/22 Oct/Nov 2013

10 Factorise completely. (a) ax + ay + bx + by Answer(a) … [2] (b) 3(x – 1)2 + (x – 1) Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 10 (a) (a + b)(x + y) 2 B1 for a(x + y) + b(x + y) or x(a + b) + y(a + b) (b) (x – 1)(3x – 2) 2 B1 for (x – 1)(3(x – 1) + 1) If B0 then SC1 for (x + a)(3x + b) where 3a + b = – 5 or ab = 2 or 3(x – 1)(x – ⅔) IGCSE – May/June 2014 0580 21 8 2 + 2 2 − 9 2

This question in 0580/21 May/June 2014

16 Factorise completely. (a) 4p2q – 6pq2 Answer(a) … [2] (b) u + 4t + ux + 4tx Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 16 (a) 2 pq (2 p − 3q ) 2 B1 for pq (4 p − 6 q ) or 2 q (2 p 2 − 3 pq ) 2 or 2 p (2 pq − 3q ) (b) (u + 4t )(1 + x ) 2 B1 for (1u + 4t ) + x (u + 4t ) or u 1( + x ) + 4t 1( + x ) 25 k 25

This question in 0580/22 May/June 2014

19 Simplify. x2 + 6 x - 7 3x + 21 Answer … [4] __________________________________________________________________________________________

4 marks

Mark scheme: 19 x − 1 final answer 4 B2 for ( x − 1)( x + 7 ) 3 or SC1 for ( x + a )( x + b ) where ab = – 7 or a + b = 6 B1 for 3( x + 7 )

This question in 0580/22 May/June 2014

14 2 8 A = 1 4 f p Work out A2 – 4A. Answer [3] f p __________________________________________________________________________________________

3 marks

Mark scheme:  4 16   12 48   8 32  14   3 M2 for   and    2 8   6 24   4 16   12 48   8 32  or M1 for   or for    6 24   4 16 

This question in 0580/21 Oct/Nov 2014

220 f(x) = 3x – 2 g(x) = , x ≠ –1 x + 1 (a) Find gf(2). Answer(a) … [2] (b) Solve g(x) = 10. Answer(b) x = … [2] (c) Simplify. f(2x) – f(x + 2) Answer(c) … [3]

7 marks

Mark scheme: 20 (a) 0.4 or 52 2 B1 for [f(2) =] 4 2 or M1 for or better (3 x − 2 ) + 1 (b) –0.8 or − 54 2 M1 for 2 = 10( x + )1 or better (c) 3 x − 6 or 3( x − 2 ) nfww 3 M2 for 3(2 x ) − 2 − (3( x + 2 ) − 2 ) or M1 for [f (2 x ) = ]3(2 x ) − 2 or [f ( x + 2 )] = 3( x + 2 ) − 2

This question in 0580/21 Oct/Nov 2014

3 7 - 2 1 22 (a) Calculate - 1 4 4 2 f fp p. Answer(a) [2] f p 5 3 (b) Calculate the inverse of 6 4 f p. Answer(b) [2] f p __________________________________________________________________________________________ Question 23 is printed on the next page.

4 marks

Mark scheme:  22 17  for a 2 × 2 matrix with 2 correct elements22 (a)   2 M1  18 7  1  4 − 3  1  a b   4 − 3  (b) 2 M1 for soi   or k    c d  2 2  − 6 5   − 6 5    or det = 2 soi

This question in 0580/21 May/June 2015

23 f(x) = 5 – 3x (a) Find f(6). Answer(a) … [1] (b) Find f(x + 2). Answer(b) … [1] (c) Find ff(x), in its simplest form. Answer(c) … [2] (d) Find f –1(x), the inverse of f(x). Answer(d) f –1(x) = … [2]

6 marks

Mark scheme: 23 (a) −13 1 (b) −3x − 1 or 5 − 3( x + 2 ) 1 (c) 9x − 10 cao 2 M1 for 5 − 3( 5 − 3x) 5 − x (d) final answer oe 2 M1 for correct first step e.g. 3 y 5 y + 3 x = 5 or = − x or y − 5 = − 3 x or 3 3 better or for interchanging x and y, e.g. x = 5 − 3 y , this does not need to be the first step

This question in 0580/21 May/June 2015

21 f(x) = x2 + 4x − 6 (a) f(x) can be written in the form (x + m)2 + n. Find the value of m and the value of n. Answer(a) m = … n = … [2] (b) Use your answer to part (a) to find the positive solution to x2 + 4x – 6 = 0. Answer(b) x = … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 21 (a) m = 2 2 B1 for m = 2 n = –10 B1 for n = –10 If 0 scored SC1 for (x + 2)2 in working or x2 + 2mx + m2 + n and equating coefficients 2m[x] = 4[x] or m2 + n = –6 (b) 1.16 or 1.16[2…] from completing 2FT FT dep on negative n square B1 for (x + their m)2 = –their n or SC1 for correct answer from using formula or for both answers 1.16 and –5.16 whatever method used

This question in 0580/22 May/June 2015

2 Factorise completely. 9x2 – 6x Answer … [2]

2 marks

Mark scheme: 2 3 x (3 x − 2 ) final answer 2 B1 for (33 x 2 − 2 x ) or x (9 x − 6 ) 2

This question in 0580/23 May/June 2015

5 Factorise 2x2 – 5x – 3. Answer … [2]

2 marks

Mark scheme: 5 ( 2 x + 1)( x − 3) 2 B1 for ( 2 x + a )( x + b ) , where ab = – 3 or a + 2b = – 5  0 1 

This question in 0580/23 May/June 2015

9 Factorise completely. (a) ax + ay + 3cx + 3cy Answer(a) … [2] (b) 3a2 – 12b2 Answer(b) … [3] __________________________________________________________________________________________

5 marks

Mark scheme: 9 (a) ( a + 3c )( x + y ) final answer 2 B1 for a ( x + y ) + 3c ( x + y ) or x ( a + 3c ) + y ( a + 3c ) (b) (3 a − 2b )( a + 2b ) final answer 3 B2 for (3 a − 2b )( a + 2b ) seen and then spoiled or (3a − 6b )( a + 2b ) or ( a − 2b )(3a + 6b ) or ( a − 2b )( a + 2b ) or B1 for 3( a 2 − 4b 2 )

This question in 0580/21 Oct/Nov 2015

15 Simplify. 2 x - 16 x 2 - 3x - 4 Answer … [4] __________________________________________________________________________________________

4 marks

Mark scheme: x + 4 15 final answer 4 B1 for ( x − 4)( x + 4) and x + 1 B2 for ( x − 4)( x + )1 or SC1 for ( x + a )( x + b ) where a + b = − 3 or ab = − 4

This question in 0580/21 Oct/Nov 2015

15 Factorise (a) 9w 2 - 100 , Answer(a) … [1] (b) mp + np - 6mq - 6nq . Answer(b) … [2]

3 marks

Mark scheme: 15 (a) (3 w + 10 )(3 w − 10 ) final answer 1 (b) (m + n )( p − 6 q ) oe final answer 2 B1 for p (m + n ) − 6 q (m + n ) oe or m ( p − 6 q ) + n ( p − 6 q ) oe 26

This question in 0580/22 Oct/Nov 2015

6 Simplify. 1 – 2u + u + 4 Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 6 5 − u final answer 2 B1 for 5 + ku or j – u , k ≠ 0 as final answer

This question in 0580/23 Oct/Nov 2015

7 Factorise completely. 2x – 4x2 Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 2 ) or x ( 2 − 4 x ) as final answer7 2 x 1( − 2 x ) final answer 2 B1 for 2( x − 2 x  360 

This question in 0580/23 Oct/Nov 2015

1 16 Make a the subject of the formula s = ut + 2 at 2. Answer a = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 2 ( s − ut )16 oe final answer 3 M1 for correctly isolating term in a 2 t M1 for correctly multiplying by 2 (or 2) M1 for correctly dividing by t2 (or t2)

This question in 0580/23 Oct/Nov 2015

22 Simplify. 4 + 10 w 8 - 50 w 2 Answer … [4] __________________________________________________________________________________________

4 marks

Mark scheme: 1 22 final answer nfww 4 B1 for 2(2 + 5w) 2 − 5 w B1 for 2(4 – 25w2) B1 for [2](2 + 5w)(2 – 5w) ALT method 4 + 10 w B3 for ( 4 + 10 w)( 2 − 5 w) or B2 for (4 + 10w)(2 – 5w) 1 2

This question in 0580/23 Oct/Nov 2015

2 Factorise 2x – 4xy. … [2]

2 marks

Mark scheme: 2 2x(1 – 2y) final answer 2 M1 for 2(x – 2xy) or x(2 – 4y) or for correct answer then spoilt 0 9

This question in 0580/22 Feb/March 2016

qx 8 y = p Write x in terms of p, q and y. x = … [2]

2 marks

Mark scheme: py 8 final answer 2 M1 for one correct step q

This question in 0580/21 May/June 2016

24 Factorise completely. (a) 2a + 4 + ap + 2p … [2] (b) 162 – 8t2 … [2]

4 marks

Mark scheme: 24 (a) ( a + 2)(2 + p ) final answer 2 B1 for 2( a + 2) + p ( a + 2) or a (2 + p ) + 2(2 + p ) (b) 2(9 + 2t )(9 − 2t ) oe 2 B1 for 2(81 − 4t 2 ) oe or (18 + 4t )(9 − 2t ) oe If 0 scored SC1 for (9 + 2t)(9 – 2t) final answer 3 3

This question in 0580/21 May/June 2016

7 Simplify. 3 3 x y + 2xy x 2 y 2 … [2]

2 marks

Mark scheme: x + 2 y x 2 y 2 27 or + 2 B1 for xy ( x + 2 y ) xy y x final answer x 2 y + 2 y 3 x 3 + 2 xy 2 or M1 for or xy 2 x 2 y pt 2t 3 p 2t + 3 p

This question in 0580/21 Oct/Nov 2016

13 Factorise completely. (a) 4p 2 - 9 … [1] (b) 2ax - 4bx - ay + 2by … [2]

3 marks

Mark scheme: 13 (a) (2 p − 3)(2 p + 3) final answer 1 (b) ( a − 2b )(2 x − y ) oe final answer 2 B1 for 2 x ( a − 2b ) − y ( a − 2b ) or a (2 x − y ) − 2b (2 x − y )

This question in 0580/21 Oct/Nov 2016

5 Simplify. 36y 5 ' 4y 2 … [2]

2 marks

Mark scheme: 5 9y 3 final answer 2 B1 for 9yk, 9 × y3 or ky3 (k ≠ 0) as final answer

This question in 0580/22 Oct/Nov 2016

13 Factorise. (a) m 3 + m … [1] (b) 25 - y 2 … [1] (c) x 2 + 3 x - 28 … [2]

4 marks

Mark scheme: 13 (a) m m 2 + 1 final answer 1 ( ) (b) ( 5 − y )( 5 + y ) final answer 1 (c) ( x − 4 )( x + 7 ) final answer 2 B1 for ( x − 4 )( x + 7 ) seen then spoiled or M1 for ( x + a )( x + b ) where ab = − 28 or a + b = 3 or for x ( x + 7) − 4( x + 7) or x ( x − 4) + 7( x − 4)

This question in 0580/22 Oct/Nov 2016

1 V = 4p2 Find V when p = 3. V = … [1]

1 marks

Mark scheme: Question Answer Mark Part marks 1 36 1

This question in 0580/23 Oct/Nov 2016

23 Simplify. 42 np - 7n 12pt - 2t + 18mp - 3 m … [4]

4 marks

Mark scheme: 23 7 n final answer 4 M1 for 7n(6p – 1) seen 2t + 3m and M2 for (2t + 3m)(6p – 1) seen or M1 for 2t(6p – 1) + 3m(6p – 1) or 6p(2t + 3m) – 1(2t + 3m)

This question in 0580/23 Oct/Nov 2016

4 2 7 - 3 - 2 3 1 - 9 25 A = B = C = D = c2 1m c4 5m c 4 5 - 1m c 0m (a) Which of these four matrix calculations is not possible? A + B 3C CB AD … [1] (b) Calculate AB. [2] f p (c) Work out B –1, the inverse of B. [2] f p (d) Explain why matrix A does not have an inverse. … [1]

6 marks

Mark scheme: 25 (a) CB 1  36 −2  (b) 2 B1 for two correct entries    18 −1  1  5 3   5 3  (c)   oe isw 2 B1 for k   seen or det = 47 soi 47  −4 7   −4 7  (d) The determinant is 0 oe 1

This question in 0580/23 Oct/Nov 2016

1 Expand the brackets and simplify. 4 (5w + 3 ) - 2 (w - 1 ) … [2]

2 marks

Mark scheme: Question Answer Marks Part Marks 1 18 w + 14 final answer 2 M1 for 20 w + 12 or − 2 w + 2 or answer 18 w + k or kw + 14

This question in 0580/22 Feb/March 2017

13 Factorise completely. (a) 15c 2 - 5c … [2] (b) 2kp - km + 6p - 3m … [2]

4 marks

Mark scheme: 3c 2 − c or c (15c − 5 )13 (a) 5c (3c − 1) final answer 2 B1 for 5 ( ) (b) ( 2 p − m )( k + 3 ) final answer 2 B1 for k ( 2 p − m ) + 3 ( 2 p − m ) or 2 p ( k + 3 ) − m ( k + 3 )

This question in 0580/22 Feb/March 2017

5 3 3 - 6 18 M = N = c1 - 2 m c4 2m Calculate (a) MN, [2] f p (b) M−1. [2] f p

4 marks

Mark scheme:  27 224  18 (a) − 5 −10  2 B1 for twwo correct elelements 1  −22 −3  −2 −3  (b) − oe issw 2 B1 for kk   oor det = −13 soi 13 − 11 5   −1 5  1

This question in 0580/22 Feb/March 2017

5 Factorise completely. 12n2 - 4mn … [2]

2 marks

Mark scheme: 5 4n(3n – m) final answer 2 B1 for 4(3n2 – mn) or n(12n – 4m) or 2n(6n – 2m) or 2(6n2 – 2mn)

This question in 0580/21 May/June 2017

6 Factorise. 14x - 21y … [1]

1 marks

Mark scheme: 6 7(2x – 3y) final answer 1

This question in 0580/22 May/June 2017

22 Factorise completely. (a) 9t 2 - u 2 … [2] (b) 2c - 4d - pc + 2pd … [2]

4 marks

Mark scheme: 22(a) ( 3t + u )( 3t − u ) final answer 2 B1 for ( at + bu )( ct + du ) final answer where ac = 9 or ad + bc = 0 or bd = – 1 22(b) ( c − 2 d )(2 − p ) or ( p − 2)(2 d − c ) 2 M1 for 2 ( c − 2 d ) − p ( c − 2 d ) final answer or c ( 2 − p ) − 2 d ( 2 − p ) or p ( 2 d − c ) − 2 ( 2 d − c ) or 2 d ( p − 2) − c ( p − 2 )

This question in 0580/22 May/June 2017

2 Factorise completely. 4x2 - 8xy … [2]

2 marks

Mark scheme: 2 4 x ( x − 2 y ) final answer 2 2 M1 for 4 x − 2 xy or x ( 4 x − 8 y ) ( ) or 2 2 x 2 − 4 xy or 2x (2x – 4y) ( )

This question in 0580/23 May/June 2017

4 Make a the subject of the formula. x = y + a a = … [2]

2 marks

Mark scheme: 4 ( x − y ) 2 oe final answer 2 M1 for x − y = a or their (x − y) squared

This question in 0580/23 May/June 2017

20 Simplify. (a) 6w0 … [1] (b) 5x3 − 3x3 … [1] (c) 3y6 # 5y-2 … [2]

4 marks

Mark scheme: 20(a) 6 1 20(b) 2x 3 final answer 1 20(c) 15y 4 final answer 2 B1 for 15 ky or ky 4 as final answer (k ≠ 0)

This question in 0580/23 May/June 2017

23 (a) Simplify. 4 (x - 6) 2 ( x - 6) … [1] (b) Expand the brackets and simplify. (x + 4) 2 + 5 (3x + 2) … [3]

4 marks

Mark scheme: 23(a) 4 ( x − 6 ) or 4x − 24 as final answer 1 23(b) x 2 + 23 x + 26 final answer 3 B2 for x 2 + 4 x + 4 x + 16 or better or B1 for 15 x + 10

This question in 0580/23 May/June 2017

5 Factorise completely. 12x 2 + 15xy - 9x … [2]

2 marks

Mark scheme: 5 3x(4x + 5y − 3) final answer 2 B1 for 3(4x2 + 5xy – 3x) or x(12x + 15y – 9) allow in working or correct answer spoiled If zero scored, SC1 for 3x(4x + 5y – 3) with only 2 correct elements in the brackets, allow in working

This question in 0580/21 Oct/Nov 2017

13 Simplify. (a) m5 2 ^ h … [1] (b) 4x 3 y # 5 x 2 y … [2]

3 marks

Mark scheme: 13(a) m10 final answer 1 13(b) 20x5y2 final answer 2 B1 for 2 out of 3 elements correct in final answer or correct answer spoiled

This question in 0580/21 Oct/Nov 2017

12 Expand the brackets and simplify. (5 - n) (3 + n) … [2]

2 marks

Mark scheme: 12 15 + 2n − n 2 final answer 2 M1 for three terms of 15 + 5n − 3n − n 2 correct

This question in 0580/22 Oct/Nov 2017

25 Factorise completely. (a) x 2 - x - 132 … [2] (b) x 3 - 4x … [2]

4 marks

Mark scheme: 25(a) ( x − 12)( x + 11) final answer 2 B1 for ( x + a )( x + b ) where ab = –132 or a + b = –1 25(b) x ( x + 2)( x − 2) final answer 2 B1 for x ( x 2 − 4) or ( x + 2)( x 2 − 2 x ) or ( x − 2)( x 2 + 2 x )

This question in 0580/22 Oct/Nov 2017

8 Expand and simplify. 6(2y - 3) - 5(y + 1) … [2]

2 marks

Mark scheme: 8 7y − 23 final answer 2 M1 for 12 y − 18 or −5 y − 5 or B1 for answer 7y − k or cy − 23 c ≠ 0

This question in 0580/21 May/June 2018

2 Expand. 7(x – 8) … [1]

1 marks

Mark scheme: 2 7x – 56 final answer 1

This question in 0580/22 May/June 2018

10 Factorise completely. xy + 2y + 3x + 6 … [2]

2 marks

Mark scheme: 10 (x + 2)(y + 3) final answer 2 B1 for y(x + 2) + 3(x + 2) or x(y + 3) + 2(y + 3)

This question in 0580/22 May/June 2018

2 Factorise. w + w3 … [1]

1 marks

Mark scheme: 2 w(1 + w 2 ) final answer 1

This question in 0580/23 May/June 2018

10 Factorise completely. 2a + 4b - ax - 2bx … [2]

2 marks

Mark scheme: 10 ( a + 2b )(2 − x ) final answer 2 M1 for 2( a + 2b ) − x ( a + 2b ) or a (2 − x ) + 2b (2 − x ) or − a ( x − 2) − 2b ( x − 2)

This question in 0580/23 May/June 2018

13 Simplify. 3 + x 9 - x 2 … [2]

2 marks

Mark scheme: 13 1 2 B1 for (3 − x )(3 + x ) or – (x – 3)(x + 3) nfww final answer 3 − x

This question in 0580/23 May/June 2018

18 Expand the brackets and simplify. 2p + 3 3p - 2 ^ ^h h … [3]

3 marks

Mark scheme: 18 6 p 2 + 5 p − 6 final answer 3 B2 for 6 p 2 + 9 p − 4 p − 6 or B1 for three correct terms

This question in 0580/23 May/June 2018

5 Expand and simplify. (3x - 7)(2x + 9) … [2]

2 marks

Mark scheme: 5 6 x 2 + 13 x − 63 final answer 2 M1 for 3 correct terms of 6 x 2 − 14 x + 27 x − 63

This question in 0580/21 Oct/Nov 2018

4 Expand. 2x 3 - x 2 ^ h … [2]

2 marks

Mark scheme: 4 6x – 2x3 final answer 2 B1 for 6x or –2x3

This question in 0580/22 Oct/Nov 2018

8 Factorise. xy + 5y + 2x + 10 … [2]

2 marks

Mark scheme: 8 (x + 5)(y + 2) final answer 2 B1 for y(x + 5) + 2(x + 5) or x(y + 2) + 5(y + 2)

This question in 0580/22 Oct/Nov 2018

5 - 3 15 M = e- 1 2o (a) Find 3M. 3M = [1] f p (b) Find M -1 . M -1 = [2] f p

3 marks

Mark scheme: 15(a)  15 −9  1 − 3 6  15(b) 1  2 3  2  2 3    oe isw B1 for k soi or det = 7 soi 7  1 5   1 5 

This question in 0580/22 Oct/Nov 2018

16 x 2 - 12x + a = ( x + b) 2 Find the value of a and the value of b. a = … b = … [3]

3 marks

Mark scheme: 16 (a = ) 36 3 B2B for a = 366 (b = ) –6 oro M1 for b == –6 oro x2 + bx + bx + b2 or better oro b2 = a

This question in 0580/22 Oct/Nov 2018

2 Factorise. y - 2y 2 … [1]

1 marks

Mark scheme: 2 y (1 − 2 y ) final answer 1

This question in 0580/23 Oct/Nov 2018

7 Simplify. 2p - q - 3q - 5p … [2]

2 marks

Mark scheme: 7 −3 p − 4 q final answer 2 B1 for –3p or –4q

This question in 0580/23 Oct/Nov 2018

11 A = r rl + r r2 Rearrange this formula to make l the subject. l = … [2]

2 marks

Mark scheme: 11 A − πr 2 A 2 M1 for A − πr 2 = πrl or πr 2 − A = − πrl or or − r oe final answer πr πr A = l + r πr

This question in 0580/23 Oct/Nov 2018

22 Simplify. 2x 2 - x - 1 2x 2 + x … [4]

4 marks

Mark scheme: 22 x − 1 1 4 B1 for x (2 x + 1) or nfww final answer x 1−x B2 for (2 x + 1)( x − 1) or B1 for 2x(x – 1) + [1](x – 1) or x(2x + 1) – [1](2x + 1) or (2 x + a )( x + b ) where ab = – 1 or a + 2b = – 1

This question in 0580/23 Oct/Nov 2018

25 Factorise completely. (a) px + py - x - y … [2] (b) 2t 2 - 98m 2 … [3]

5 marks

Mark scheme: 25(a) ( x + y )( p − 1) final answer 2 M1 for p ( x + y ) − ( x + y ) or x ( p − 1) + y ( p − 1) 25(b) 2(t + 7 m )(t − 7 m ) final answer 3 M2 for (2t + 14 m )( t − 7 m ) or (t + 7 m )(2t − 14 m ) or correct answer seen or M1 for 2(t 2 − 49 m 2 ) or (t + 7 m )( t − 7 m ) or 2(t + 7)( t − 7)

This question in 0580/23 Oct/Nov 2018

13 Factorise. (a) 7k 2 - 15 k … [1] (b) 12 (m + p ) + 8 (m + p ) 2 … [2]

3 marks

Mark scheme: 13(a) k ( 7 k − 15 ) final answer 1 13(b) 4 ( m + p )( 3 + 2 m + 2 p ) 2 B1 for (m + p)(12 + 8(m + p)) or (m + p)(12 + 8m + 8p) final answer or (4m + 4p)(3 + 2m + 2p) or (2m + 2p)(6 + 4m + 4p) or 2(2m + 2p)(3 + 2m + 2p) or 2(m + p)(6 + 4m + 4p)

This question in 0580/22 Feb/March 2019

19 Simplify. 2 ab - b a 2 - b 2 … [3]

3 marks

Mark scheme: 19 b 3 B1 for b ( a − b ) final answer a + b B1 for ( a + b )( a − b )

This question in 0580/22 Feb/March 2019

2 - 1 1 6 20 (a) Work out e4 3eo- 5 4o. [2] f p 3 - 1 (b) Find the value of x when the determinant of is 5. e- 7 xo x = … [2]

4 marks

Mark scheme: 20(a)  7 8  2 B1 for 2 correct elements    −11 36  20(b) 4 2 M1 for 3 x −−( 1) × ( −7 ) = 5 or better

This question in 0580/22 Feb/March 2019

2 Factorise 5y - 6py. … [1]

1 marks

Mark scheme: 2 y (5 − 6 p ) final answer 1

This question in 0580/21 May/June 2019

13 x 2 + 4x - 9 = (x + a) 2 + b Find the value of a and the value of b. a = … b = … [3]

3 marks

Mark scheme: 13 [a = ] 2 3 B2 for either correct or (x + 2)2 – 13 [b = ] – 13 OR M1 for 2a = 4 soi M1 for a2+ b = – 9 soi OR M1 for x2 + ax + ax + a2 [+b] or better

This question in 0580/21 May/June 2019

15 Expand and simplify. (x + 1)(x + 2) + 2x (x - 3) … [3]

3 marks

Mark scheme: 15 3 x 2 − 3 x + 2 final answer 3 B2 for x 2 + 2 x + x + 2 + 2 x 2 − 6 x oe or B1 for 3 correct terms of x 2 + 2 x + x + 2 oe

This question in 0580/21 May/June 2019

17 (a) Factorise p 2 - q 2 . … [1] (b) p 2 - q 2 = 7 and p - q = 2 . Find the value of p + q. … [2]

3 marks

Mark scheme: 17(a) ( p − q )( p + q ) final answer 1 17(b) 7 2 M1 for 2 × (p + q) = 7 oe 2 2 2 2 2 or for ( 2 + q ) − q = 7 or p − ( p − 2 ) = 7

This question in 0580/21 May/June 2019

3 1 23 P = e2 4o (a) Find P2. [2] f p (b) Find P–1. [2] f p

4 marks

Mark scheme: 23(a)  11 7  2 B1 for 2 or 3 correct elements    14 18  23(b) 1  4 − 1  2  4 − 1    oe isw B1 for k   or for det = 10 soi 10  − 2 3   − 2 3 

This question in 0580/21 May/June 2019

19 Rearrange this formula to make m the subject. k + m P = m … [4]

4 marks

Mark scheme: 19 k 4 k m = final answer B3 for final answer P − 1 P − 1 OR M1 for multiplying or dividing by m correctly M1 for term(s) in m on one side correctly and terms not in m on the other side correctly M1 for correctly factorising m with a 2-term bracket oe M1 for correct division by their 2-term bracket with m as the subject To a maximum of M3 for an incorrect answer

This question in 0580/22 May/June 2019

2 7 3 4 22 A = B = e1 3o e0 1o (a) Calculate AB. [2] f p (b) Find A -1 , the inverse of A. [2] f p

4 marks

Mark scheme: 22(a)  6 15  2 B1 for 2 correct elements    3 7  22(b)  − 3 7  2  3 −7    oe isw B1 for k   soi or det = − 1 soi  1 − 2   −1 2 

This question in 0580/22 May/June 2019

2 Factorise 2x 2 - x . … [1]

1 marks

Mark scheme: 2 x(2x – 1) 1

This question in 0580/23 May/June 2019

11 Complete this statement with an expression in terms of m. 18m 3 + 9m 2 + 14m + 7 = (9m 2 + 7) ( … ) [2]

2 marks

Mark scheme: 11 2m + 1 2 B1 for 2m + c or km + 1 (k ≠ 0)

This question in 0580/23 May/June 2019

3 4 1 4 21 A = B = e5 0o e- 3 2o Find (a) 5A, [1] f p (b) A + B , [1] f p (c) AB. [2] f p

4 marks

Mark scheme: 21(a)  15 20  1    25 0  21(b)  4 8  1    2 2  21(c)  −9 20  2 B1 for two correct elements    5 20 

This question in 0580/23 May/June 2019

2 Factorise 5p + pt. … [1]

1 marks

Mark scheme: 2 p(5 + t) final answer 1

This question in 0580/21 Oct/Nov 2019

5 Simplify 5c - d - 3d - 2 c . … [2]

2 marks

Mark scheme: 5 3c – 4d final answer 2 B1 for 3c + kd or kc – 4d

This question in 0580/21 Oct/Nov 2019

7 Simplify 2x 3 # 3x 2 . … [2]

2 marks

Mark scheme: 7 6x 5 final answer 2 B1 for kx 5 or 6 x k

This question in 0580/21 Oct/Nov 2019

11 P = 2r + r r Rearrange the formula to write r in terms of P and r. r = … [2]

2 marks

Mark scheme: 11 P 2 M1 for P = r (2 + π) 2 + π

This question in 0580/21 Oct/Nov 2019

3 Expand. a ( a 3 + 3) … [1]

1 marks

Mark scheme: 3 a4 + 3a final answer 1

This question in 0580/22 Oct/Nov 2019

20 (a) Factorise. 18y - 3ay + 12x - 2ax … [2] (b) Factorise. 3x 2 - 48y 2 … [3]

5 marks

Mark scheme: 20(a) (3y + 2x)(6 – a) oe final answer 2 M1 for 3y (6 – a) + 2x(6 – a) oe or 6(2x + 3y) – a(2x + 3y) oe 20(b) 3(x + 4y)(x – 4y) final answer 3 M2 for (3x + 12y)(x – 4y) or (3x – 12y)(x + 4y) or M1 for 3(x2 – 16y2) or for (x + 4y)(x – 4y)

This question in 0580/22 Oct/Nov 2019

3 2 - 2 5 22 A = B = C = (- 1 k) e- 5 0o e 4 1o (a) Find AB. [2] f p (b) CA = (- 13 - 2 ) Find the value of k. k = … [2] (c) Find A-1. [2] f p

6 marks

Mark scheme: 22(a)  2 17  2 B1 for 2 correct elements    10 − 25  22(b) 2 2 M1 for –3 –5k = –13 oe 22(c) 1  0 − 2  2  0 − 2    oe isw M1 for k   or for det = 10 or soi 10  5 3   5 3 

This question in 0580/22 Oct/Nov 2019

6 Expand and simplify (x + 3)(x + 5). … [2]

2 marks

Mark scheme: 6 x2 + 8x + 15 final answer 2 M1 for three terms correct from x2 + 3x + 5x + 15

This question in 0580/23 Oct/Nov 2019

11 Factorise. (a) 12x + 15 … [1] (b) xy - 2x + 3y - 6 … [2]

3 marks

Mark scheme: 11(a) 3(4x + 5) final answer 1 11(b) (x + 3)(y – 2) final answer 2 B1 for y(x + 3) – 2(x + 3) or x(y – 2) + 3(y – 2) or correct answer seen then spoilt

This question in 0580/23 Oct/Nov 2019

9 (a) Factorise completely. 3x 2 - 12xy … [2] (b) Expand and simplify. ( m - 3)( m + 2) … [2]

4 marks

Mark scheme: 9(a) 3 x ( x − 4 y ) final answer 2 B1 for 3( x 2 − 4 xy ) or x (3 x − 12 y ) 9(b) m 2 − m − 6 final answer 2 M1 for 3 terms from m 2, − 3m, + 2 m, − 6

This question in 0580/22 Feb/March 2020

20 x 2 - 12x + a = ( x + b) 2 Find the value of a and the value of b. a = … b = … [2]

2 marks

Mark scheme: 20 [a =] 36 2 B1 for each [b =] − 6 or SC1 for correct answers reversed

This question in 0580/22 Feb/March 2020

9 Factorise completely. (a) 21a 2 + 28ab … [2] (b) 20x 2 - 45y 2 … [3]

5 marks

Mark scheme: 9(a) 7a(3a + 4b) final answer 2 B1 for partial factorisation 7(3a2 + 4ab) or a(21a + 28b) 9(b) 5(2x + 3y)(2x – 3y) final 3 B2 for (2x + 3y)(2x – 3y) answer or (10x + 15y)(2x – 3y) or (2x + 3y)(10x – 15y) or B1 for 5(4x2 – 9y2)

This question in 0580/21 May/June 2020

21 Expand and simplify ( x + 3)( x - 5)( 3x - 1) . … [3] Question 22 is printed on the next page.

3 marks

Mark scheme: 21 3x3 – 7x2 – 43x + 15 3 B2 for correct expansion and simplification of two of the brackets or B1 for correct expansion of two brackets with at least 3 terms correct

This question in 0580/21 May/June 2020

10 Simplify. p 4pq 2q # t … [2]

2 marks

Mark scheme: 10 2p 2 2 B1 for correct unsimplified answer t

This question in 0580/22 May/June 2020

19 Make y the subject of the formula. h 2 = x 2 + 2y 2 y = … [3]

3 marks

Mark scheme: 19 2 2 3 M1 for correct rearrangement for y or y2 term h − x [±] M1 for correct square root 2 M1 for correct division by 2 or 2

This question in 0580/22 May/June 2020

25 Simplify. 2x 2 + x - 15 ax + 3a - 2bx - 6b … [5]

5 marks

Mark scheme: 25 2 x − 5 5 B2 for (2x – 5)(x + 3) final answer or B1 for (2x + p)(x + q) where pq = –15 or a − 2b p + 2q = 1 B2 for (x + 3)(a – 2b) or B1 for x(a – 2b) + 3(a – 2b) or a(x + 3) – 2b(x + 3)

This question in 0580/22 May/June 2020

18 (a) Write x 2 - 18 x - 27 in the form ( x + k) 2 + h . … [2] (b) Use your answer to part (a) to solve the equation x 2 - 18x - 27 = 0 . x = … or x = … [2]

4 marks

Mark scheme: 18(a) ( x − 9 )2 − 108 2 B1 for ( x + h ) 2 − 108 or ( x − 9 ) 2 + h or k = − 9 18(b) 19.4 or 19.39… 2 M1FT x − their 9 = ± their108 −1.39 or −1.392… A1 for 9 ± 108 or 9 ± 6 3

This question in 0580/23 May/June 2020

1 Simplify. 3a + 7b - 4a + b … [2]

2 marks

Mark scheme: Question Answer Marks Partial Marks 1 –a + 8b final answer 2 B1 for –a or [+]8b in final answer or for –a + 8b spoilt

This question in 0580/21 Oct/Nov 2020

16 Factorise 6x 2 + 7x - 20 . … [2]

2 marks

Mark scheme: 16 (3x – 4)(2x + 5) final answer 2 B1 for (ax + b)(cx + d) where ac = 6 and ad + bc = 7 or bd = – 20

This question in 0580/21 Oct/Nov 2020

6 Factorise completely. 4 - 8 x … [1]

1 marks

Mark scheme: 6 4(1 – 2x) 1

This question in 0580/22 Oct/Nov 2020

26 Simplify. ux - 2u - x + 2 u 2 - 1 … [4]

4 marks

Mark scheme: 26 x − 2 4 B2 for ( x − 2)(u − 1) oe final answer u + 1 or B1 for u ( x − 2) − ( x − 2) or x (u − 1) − 2(u − 1) B1 for (u − 1)( u + 1)

This question in 0580/22 Oct/Nov 2020

x15 m = 2p + y Make x the subject of this formula. x = … [3]

3 marks

Mark scheme: 15 [x =] y(m – 2p)2 nfww 3 M1 for subtract 2p or their term in p to isolate a term in x or [x =] y(m2 – 4mp + 4p2) final answer M1 for squaring M1 for multiplying by their term in y Maximum of 2 marks for an incorrect answer

This question in 0580/23 Oct/Nov 2020

20 Factorise. 3x + 8y - 6ax - 16ay … [2]

2 marks

Mark scheme: 20 (3x + 8y)(1 – 2a) 2 M1 for 3x(1 – 2a) + 8y(1 – 2a) or 3x + 8y – 2a(3x + 8y) or better

This question in 0580/23 Oct/Nov 2020

24 Simplify. x 2 - 25 x 2 - 17x + 60 … [4] Question 25 is printed on the next page.

4 marks

Mark scheme: 24 x + 5 4 B1 for (x + 5) (x – 5) nfww final answer B2 for (x – 12) (x – 5) x − 12 or B1 for x(x – 5) – 12 (x – 5) or x(x – 12) – 5(x – 12) or for (x + a)(x + b) where ab = –60 or a + b = –17

This question in 0580/23 Oct/Nov 2020

15 Make h the subject of the formula 2mh = g ( 1 - h) . h = … [4]

4 marks

Mark scheme: 15 g 4 M1 for expanding brackets or ÷g final answer M1 for isolating terms in h 2 m + g M1 for factorising M1 for dividing by bracket to isolate h Incorrect/unsimplified final answer scores max 3 marks

This question in 0580/21 May/June 2021

20 Expand and simplify. ( x - 2)( 2x + 5)( x + 3) … [3]

3 marks

Mark scheme: 20 2 x 3 + 7 x 2 − 7 x − 30 final answer 3 B2 for unsimplified expansion with at most one error or for simplified four-term expression of correct form with three terms correct or B1 for correct expansion of two brackets with at least three terms out of four correct

This question in 0580/22 May/June 2021

13 Expand and simplify. 6 ( t - q ) - 2 ( t - 3q) … [2]

2 marks

Mark scheme: 13 4t final answer 2 B1 for 6t – 6q or – 2t + 6q or 2t – 6q or for 4t or 0q in the final answer

This question in 0580/23 May/June 2021

13 (a) Write 243 # 27 2 n as a single power of 3 in terms of n. … [2] (b) k = 2 # 32 # p 3 , where p is a prime number greater than 3. Write 6k 2as a product of prime factors in terms of p. … [2]

4 marks

Mark scheme: 13(a) 36n + 5 final answer 2 B1 for 35 or (33)2n or better or answer 6n + 5 13(b) 23 × 35 × p6 final answer 2 B1 for two parts correct or 2 × 3 × 2 × 32 × p3 × 2 × 32 × p3 or 1944p6 or k2 = 22 × 34 × p6

This question in 0580/22 Oct/Nov 2021

21 Expand and simplify. ( x - 3) 2 ( 2x + 5) … [3]

3 marks

Mark scheme: 21 2 x 3 − 7 x 2 − 12 x + 45 3 B2 for unsimplified expansion of the three brackets with at most one error final answer or for simplified four-term expression of correct form with three terms correct or B1 for correct expansion of two of the given brackets with at least three terms out of four correct

This question in 0580/22 Oct/Nov 2021

11 P = M ( g 2 + h 2 ) (a) Find the value of P when M = 100, g = 3 and h = 4.5 . P = … [2] (b) Rearrange the formula to write g in terms of P, M and h. g = … [3]

5 marks

Mark scheme: 11(a) 2925 2 M1 for 100(32 + 4.52) or B1 for 29.25 seen 11(b) P 2 3 M1 for correct division by M [±] − h M1 for correct re-arrangement to isolate g M or g2 P − Mh 2 M1 for correct square root of two term or [±] final answer M expression Max 2 marks for an incorrect answer

This question in 0580/23 Oct/Nov 2021

25 Simplify. 3x 2 - 18x ax - 6a + 2cx - 12c … [4]

4 marks

Mark scheme: 25 3 x 4 B1 for 3x(x – 6) final answer B2 for ( x − 6)( a + 2c ) a + 2c or B1 for a ( x − 6) + 2c ( x − 6) or x ( a + 2c ) − 6( a + 2 c )

This question in 0580/23 Oct/Nov 2021

9 Factorise completely. 12a 3 - 21a … [2]

2 marks

Mark scheme: 9 2 2 3 2 3a 4 a − 7 final answer B1 for 3 4 a − 7 a or a 12 a − 21 ( ) ( ) ( ) or for 3a seen then spoilt ( 4 a 2 − 7 )

This question in 0580/22 Feb/March 2022

1 28 s = at 2 (a) Work out the value of s when a = 0.9 and t = 4. s = … [1] (b) Rearrange the formula to find t in terms of s and a. t = … [2]

3 marks

Mark scheme: 8(a) 7.2 oe 1 8(b) 2s 2 s 1 2 [±] final answer M1 for  t or 2s = at2 or better a a 2 2 xy  y 2 or y 14 x  7 y  9 7 y  2 x  y  final answer 2 B1 for 7   or 7 y  2 x  y  seen then spoilt

This question in 0580/21 May/June 2022

21 Factorise completely. 1 – q – a + aq … [2]

2 marks

Mark scheme: 21 (1 – q)(1 – a) or (a – 1)( q – 1) 2 B1 for 1 – q – a(1 – q) or 1 – a – q(1 – a) or final answer better or correct answer seen and spoilt

This question in 0580/21 May/June 2022

23 x 2 + 8x + 10 = ( x + p) 2 + q (a) Find the value of p and the value of q. p = … q = … [2] (b) Solve. x 2 + 8x + 10 = 30 x = … or x = … [2]

4 marks

Mark scheme: 23(a) [p = ] 4 2 B1 for one correct [q = ] –6 2 or  x  4   6 or x2 + px + px + p2 [+ q] 23(b) –10 and 2 2 2 M1 for  x  4   36 or  x  their 4  2  30  their  6  or for correct method to solve quadratic e.g.  x  10  x  2 

This question in 0580/21 May/June 2022

20 Factorise completely. (a) 2m + 3p - 8km - 12kp … [2] (b) 5x 2 - 20y 2 … [3]

5 marks

Mark scheme: 20(a)  2 m  3 p 1  4 k  final 2 B1 for 2 m  3 p  4 k  2 m  3 p  or better answer or 2 m 1  4 k   3 p 1  4 k  or correct answer seen and spoilt 20(b) 5  x  2 y  x  2 y  final 3 B2 for (5x – 10y)(x + 2y) or (x – 2y)(5x + 10y) or correct answer seen then spoilt answer or B1 for 5 x 2  4 y 2   or for  x  2 y  x  2 y 

This question in 0580/22 May/June 2022

13 Factorise completely. (a) 18px - 27p … [2] (b) mt - n - m + nt … [2]

4 marks

Mark scheme: 13(a) 9p(2x – 3) final answer 2 B1 for 9(2px – 3p) or p(18x – 27) or 3p(6x – 9) or 9p(2x – 3) seen and spoilt 13(b) (m + n)(t – 1) final answer 2 B1 for m(t – 1) + n(t – 1) or t(m + n) – [1](m + n) or correct answer seen and spoilt

This question in 0580/23 May/June 2022

19 Expand and simplify. ( 2x + 3)( x - 2) 2 … [3]

3 marks

Mark scheme: 19 2 x 3 − 5 x 2 − 4 x + 12 final answer 3 B2 for correct expansion of the three brackets unsimplified or for simplified four- term expression of correct form with three terms correct or B1 for correct expansion of two of the three given brackets with at least three terms out of four correct

This question in 0580/21 Oct/Nov 2022

20 Factorise completely. (a) 1+ x - y - xy … [2] (b) 2x 3 - 18xy 2 … [3]

5 marks

Mark scheme: 20(a) (1 + x )(1 − y ) final answer 2 B1 for 1 + x − y (1 + x ) or 1 – y + x(1 – y) or correctly x 2 − 9 y 220(b) 2 x ( x + 3 y )( x − 3 y ) final answer 3 B2 for 2 x ( ) factorising into two brackets e.g. 2 x 2 + 6 xy x 2 − 3 xy 2 x + 6 y ) ( )( x − 3 y ) , ( )( or B1 for 2 x 3 − 9 xy 2 or x 2 x 2 − 18 y 2 ( ) ( ) or for ( x + 3 y )( x − 3 y )

This question in 0580/21 Oct/Nov 2022

2 Simplify. y # 27 - y # 77 … [1]

1 marks

Mark scheme: 2 –50y 1

This question in 0580/22 Oct/Nov 2022

4 Expand. x ( 3 + x 2 ) … [2]

2 marks

Mark scheme: 4 3x + x3 final answer 2 B1 for one correct term from two in final answer or for correct answer then spoilt

This question in 0580/22 Oct/Nov 2022

22 (a) Expand and simplify. ( 2x - 1)( x + 4)( x - 3) … [3] (b) Write as a single fraction in its simplest form. 4 2 x 2 + 14 x ' 2x - 3 2x 2 + 11x - 21 … [4]

7 marks

Mark scheme: 22(a) 2x3 + x2 – 25x + 12 final answer 3 B2 for correct unsimplified expanded expression or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of 2 brackets with at least 3 terms out of 4 correct 22(b) 2 4  4 2 x 2 + 11x − 21 final answer M1 for  oe soi   2 x  2 x − 3  2 x + 14 x B1 for (x + 7)(2x – 3) oe factorised B1 for 2x(x + 7) oe factorised

This question in 0580/23 Oct/Nov 2022

5 Factorise completely. 8g - 2g 2 … [2]

2 marks

Mark scheme: 5 2g(4 – g) final answer 2 B1 for 2(4g – g2) or for g(8 – 2g) or for 2g(4 – g) seen then spoiled

This question in 0580/22 Feb/March 2023

5 Factorise completely. (a) 42mk - 35m … [2] (b) h 2 - 144 … [1]

3 marks

Mark scheme: 5(a) 7m(6k – 5) final answer 2 B1 for 7(6mk – 5m) or m(42k – 35) as final answer or 7m(6k – 5) seen and then spoiled 5(b) (h + 12)(h – 12) final answer 1

This question in 0580/21 Oct/Nov 2023

10 Expand and simplify. 2 ( t + w) + 3 ( w - t) … [2]

2 marks

Mark scheme: 10 5w – t final answer 2 B1 for 2t + 2w or 3w – 3t or for 5w – t seen then spoiled or for 5w or – t in the final answer

This question in 0580/22 Oct/Nov 2023

24 Simplify. ax - 2a - x + 2 a 2 - 1 … [4]

4 marks

Mark scheme: 24 x – 2 4 B2 for ( x − 2 )( a − 1) final answer a + 1 or M1 for a ( x − 2 ) − ( x − 2 ) or x ( a − 1) − 2 ( a − 1) B1 for ( a − 1)( a + 1)

This question in 0580/22 Oct/Nov 2023

20 ( x + a)( x + 2)( 2x + 3) is equivalent to 2x 3 + bx 2 + cx - 18 . Find the value of each of a, b and c. a = … b = … c = … [3]

3 marks

Mark scheme: 20 [a =] – 3 3 B1 for a = – 3 [b =] 1 B1FT for b = 7 + 2  their a [c =] – 15 B1FT for c = 6 + 7  their a If B0 scored B1 for correct expansion of a pair of brackets or of three brackets x 2 + ax + 2 x + 2a 2 x + 3 or ( ) or 2 x 2 + 4 x + 3 x + 6  x + a ( ) 2 x3 + ( 2a + 7 ) x 2 + ( 7a + 6 ) x + 6a oe or for b = 7 + 2a or for c = 6 + 7a

This question in 0580/23 Oct/Nov 2023

3 Simplify 4m + 7k - m + 3k . … [2]

2 marks

Mark scheme: 3 3m + 10k final answer 2 B1 for 3m or 10k in final answer or for 3m + 10k seen and spoilt

This question in 0580/22 Feb/March 2024

18 Find the highest common factor (HCF) of 28x 5 and 98x 3. … [2]

2 marks

Mark scheme: 18 14x 3 2 B1 for 14 kx or 7x 3 or 2x 3

This question in 0580/22 Feb/March 2024

24 x 2 - 16 x + a can be written in the form ( x + b) 2 . Find the value of a and the value of b. a = … b = … [2] Questions 25 and 26 are printed on the next page.

2 marks

Mark scheme: 24 [a =] 64 2 B1 for each [b =] −8 or for both (x – 8)2 and x2 – 16x + 64

This question in 0580/22 Feb/March 2024

19 Factorise completely. (a) 12m 2 - 75t 2 … [3] (b) xy + 15 + 3y + 5x … [2]

5 marks

Mark scheme: 19(a) 3(2m + 5t)(2m – 5t) final answer 3 B2 for (6m + 15t)(2m – 5t) or (2m + 5t)(6m – 15t) or B1 for 3(4m2 – 25t2) or (2m + 5t)(2m – 5t) 19(b) (x + 3)(y + 5) final answer 2 B1 for x(y + 5) + 3(y + 5) or y(x + 3) + 5(x + 3)

This question in 0580/21 May/June 2024

3 Simplify. 7x - 8y - x - y … [2]

2 marks

Mark scheme: 3 6x – 9y or 3(2x – 3y) final answer 2 B1 for 6x or – 9y in final answer or 6x – 9y seen then spoilt

This question in 0580/23 May/June 2024

6 Factorise completely. 4x 2 y - 5xy 2 … [2]

2 marks

Mark scheme: 6 xy(4x – 5y) final answer 2 B1 for y(4x2 – 5xy) or x(4xy – 5y2) or xy(4x – 5y) seen then spoilt

This question in 0580/23 May/June 2024

16 A = rr 2 + rdh Rearrange the formula to make h the subject. h = … [2]

2 marks

Mark scheme: 16 A  r 2 2 2 A r 2 oe final answer M1 for A  r  dh or   h d d d A 2 or  r  dh

This question in 0580/23 May/June 2024

7 Factorise. 28x - 35 … [1]

1 marks

Mark scheme: 7 7(4x – 5) final answer 1

This question in 0580/21 Oct/Nov 2024

22 The graph of y = ( x + 2)( x - 1) 2 is shown on the grid. y 4 3 2 1 x -2 -1 0 1 2 -1 (a) Show that y = ( x + 2 )( x - 1 ) 2 can be written as y = x 3 - 3x + 2 . [2] (b) By drawing a suitable straight line, solve the equation 2x 3 - 5x = 0 . x = … or x = … or x = … [4] Question 23 is printed on the next page.

6 marks

Mark scheme: 22(a) x2 – x – x + 1 M1 or x2 + 2x– x – 2 A correct unsimplified expansion A1 e.g. x3 + 2x2 –x2 –2x – x2 –2x + x + 2 oe leading to [y = ] x3 – 3x + 2 22(b) y = 2 – 0.5x ruled B2 B1 for [y =] 2 – 0.5x soi or for y = 2 – kx drawn or for y = k – 0.5x drawn –1.5 to –1.6 B2 B1 for two correct values 0 1.5 to 1.6

This question in 0580/21 Oct/Nov 2024

23 ( x - 5) 2 + k = x 2 - px - 21 Find the value of p and the value of k. p = … k = … [2]

2 marks

Mark scheme: 23 [p = ] 10 2 B1 for each [k = ] –46

This question in 0580/21 Oct/Nov 2024

11 Factorise fully. (a) 24x 2 - 9 xy … [2] (b) 63x 2 - 28y 2 … [3]

5 marks

Mark scheme: 11(a) 3x(8x – 3y) final answer 2 B1 for 3(8x2 – 3xy) or x(24x – 9y) or 3x(8x – 3y) seen then spoilt 11(b) 7(3x + 2y)(3x – 2y) final answer 3 B2 for (21x + 14y)(3x – 2y) or (3x + 2y)(21x – 14y) or 7(3x + 2y)(3x – 2y) seen then spoilt or M1 for 7(9x2 – 4y2) or [...](3x + 2y)(3x – 2y)

This question in 0580/22 Oct/Nov 2024

21 Expand and simplify. ( x + 3)( x + 5)( 2x + 1) … [3]

3 marks

Mark scheme: 21 2x3 + 17x2 + 38x + 15 3 B2 for correct expansion of the three brackets final answer unsimplified or for simplified four-term expression of correct form with three terms correct or B1 for correct expansion of two of the given brackets with at least three terms out of four correct

This question in 0580/22 Oct/Nov 2024

12 (a) Write as a single fraction in its simplest form. x 3 x x + 2 + - 4 8 12 … [3] (b) Factorise. 3 x ( a + 4y) - ay - 4y 2 … [1]

4 marks

Mark scheme: 12(a) 13 x − 4 3 13 x + 4 final answer SC2 for final answer 24 24 6 x + 9 x − 2 x − 4 or M2 for oe or 24 better 6 x + 3 ( 3 x ) − 2 ( x + 2 ) or M1 for oe 24 12(b) (3x – y)(a + 4y) final answer 1

This question in 0580/22 Feb/March 2025

1 Simplify. 7 c - 5 d + c + 3d … [2]

2 marks

Mark scheme: Question Answer Marks Partial Marks 1 8c – 2d final answer 2 B1 for answer 8c – kd or kc – 2d or for correct answer seen and spoilt

This question in 0580/21 May/June 2025

16 Expand and simplify. ( x + 4)( x - 3)( 3x + 2 ) … [3]

3 marks

Mark scheme: 16 3x3 + 5x2 – 34x – 24 final answer 3 B2 for correct expansion unsimplified or simplified four-term expression of correct form with 3 terms correct or B1 for one pair of brackets expanded with at least 3 terms out of 4 correct

This question in 0580/21 May/June 2025

22 A curve has equation y = x n + qx 2 + 9x . dy 2 = 3 x - 12 x + 9 dx (a) Find the value of n, and the value of q. n = … q = … [2] (b) Work out the coordinates of the turning points of the curve. ( … , … ) and ( … , … ) [4]

6 marks

Mark scheme: 22(a) [n =] 3, [q =] – 6 2 B1 for each correct value 22(b) (1, 4) and (3, 0) 4 B3 for (1, 4) or (3, 0) or for two correct values of x or M2 for [3](x – 1)(x – 3) [ = 0] oe −−( 12 )  ( −12 ) 2 −4 3 9 or x = oe 2  3 d y or M1 for writing = 0 d x or for 3 x 2 − 12 x + 9 = 0

This question in 0580/21 May/June 2025

23 Simplify. 2x 2 + 10 x x 2 - 25 … [3]

3 marks

Mark scheme: 23 2 x 3 B1 for 2x(x + 5) final answer x − 5 B1 for (x – 5)(x + 5)

This question in 0580/21 May/June 2025

6 Solve. (a) 8x + 7 = 39 x = … [2] (b) 2 ( 5y - 1 ) = 24 y = … [3]

5 marks

Mark scheme: 6(a) 4 2 M1 for 8x = 39 – 7 or better 6(b) 13 3 M1 for correct first step e.g. 2.6 or oe 5 24 5y – 1 = or 10y – 2 = 24 or better 2 M1 for correctly isolating terms in y FT their first step e.g. 5y = 12 + 1 or 10y = 24 + 2

This question in 0580/22 May/June 2025

19 Write as a single fraction in its simplest form. 5a 3 b (a) # 6 a … [2] p 3t (b) + 2 4 … [2] 2 3 (c) - x - 2 x + 1 … [3]

7 marks

Mark scheme: 19(a) 5b 2 15 ab cao final answer B1 for or better seen 2 6 a 19(b) 2 p + 3t 2 M1 for adding two correct fractions with a cao final answer 4 p 6t 4 common denominator e.g. + 8 8 19(c) 8 − x 8 − x 3 B1 for 2(x + 1) – 3(x – 2) or better isw or 2 −−x 2 ( x − 2 )( x + 1) x B1 for common denominator (x – 2)(x + 1) oe isw cao final answer

This question in 0580/22 May/June 2025

24 The line y = 7x + 3 intersects the curve y = x 2 + 5x - 12 at the points A and B. Find the coordinates of A and B. A ( … , … ) B ( … , … ) [5]

5 marks

Mark scheme: 24 (5, 38) and (–3, –18) 5 B4 for one correct coordinate or for x = 5 and x = –3 OR M2 for x2 – 2x – 15 [= 0] or y2 – 20y – 684 [= 0] or M1 for 7x + 3 = x2 + 5x – 12 oe  y − 3  2  y − 3  or y = + 5 − 12      7   7  M1 for correct method to solve their three- term quadratic (x – 5)(x + 3) −−( 2 )  ( −2 ) 2 −−4 1 15 oe 2  1 If B0 scored and at least 2 method marks scored, SC1 for correct substitution of both of their x values or their y values into y = 7x + 3 or y = x2 + 5x – 12

This question in 0580/22 May/June 2025

19 Simplify. 7x - x 2 49 - x 2 … [3]

3 marks

Mark scheme: 19 x 3 B1 for x(7 – x) final answer B1 for (7 – x)(7 + x) 7 + x

This question in 0580/23 May/June 2025

7 Simplify. p 2 (a) ' t t … [2] 3x x - 1 (b) - 4 2 … [2]

4 marks

Mark scheme: 7(a) p 2 p t final answer M1 for  or for cross cancelling of t 2 t 2 7(b) x + 2 x 1 2 3 x − 2 ( x − 1) or + final answer M1 for oe 4 4 2 4 x − 2 If 0 scored, SC1 for final answer 4 oe

This question in 0580/21 Oct/Nov 2025

17 I = M ( k2 + c 2 ) (a) Find the value of I when M = 7, k = 3 and c = 2. I = … [2] (b) Rearrange the formula to write k in terms of I, M and c. k = … [3]

5 marks

Mark scheme: 17(a) 91 2 M1 for 7(32 + 22) oe 17(b) 2 3 M1 for correctly dividing by M I 2 I − Mc  − c or  M1 for correctly isolating term in k2 M M M1 for correctly taking square root final answer Maximum M2 if answer is incorrect

This question in 0580/21 Oct/Nov 2025

18 f ( )x = 2x + 5 f ( x) f ( x) - ff ( x) = ax 2 + bc + c Find the value of a, the value of b and the value of c. a = … b = … c = … [4]

4 marks

Mark scheme: 18 [a =] 4 4 B3 for two correct nfww [b =] 16 [c =] 10 OR B2 for 4 x 2 + 20 x + 25 or B1 for three terms correct from 4 x 2 + 10 x + 10 x + 25 M1 for 2(2x + 5) + 5 soi

This question in 0580/21 Oct/Nov 2025

25 Simplify. 10 ax + 6bx - 25a - 15 b 4 x 2 - 25 … [4]

4 marks

Mark scheme: 25 5a + 3b 4 B2 for (5a + 3b )(2 x − 5) final answer 2 x + 5 or B1 for 5a (2 x − 5) + 3b(2 x − 5) or for 2 x (5a + 3b ) − 5(5a + 3b ) B1 for (2 x + 5)(2 x − 5)

This question in 0580/21 Oct/Nov 2025

10 b = dm + 2mk (a) d = 3.14 , m = 7.92 and k = 10.16 . By rounding each value correct to 1 significant figure, work out an estimate for b. … [3] (b) Rearrange the formula to make m the subject. m = … [2]

5 marks

Mark scheme: 10(a) 184 3 B1 for 3, 8 and 10 with 3, 8 and 10 shown M1 for 3 × 8 + 2 × 8 × 10 or for correct substitution of unrounded, truncated or incorrectly rounded values 10(b) b 2 M1 for [b =] m(d + 2k) [m =] final answer d + 2k

This question in 0580/22 Oct/Nov 2025

15 Factorise. (a) x 2 - 64 … [1] (b) 5 x ( x - 2y) + 6 ( x - 2y) 2 … [2]

3 marks

Mark scheme: 15(a) (x + 8)(x – 8) final answer 1 15(b) (x – 2y)(11x – 12y) oe final answer 2 M1 for (x – 2y)(5x + 6(x – 2y)) or for (x + ay)(11x + by) where ab = 24 or 11a + b = – 34 or for correct answer seen and spoilt. If 0 scored, SC1 for 11x2 – 34xy + 24y2

This question in 0580/22 Oct/Nov 2025

15 Factorise. (a) x 2 - 7x + 12 … [2] (b) 5 x + 10 y + 6 ny + 3 nx … [2]

4 marks

Mark scheme: 15(a) (x – 3)(x – 4) final answer 2 B1 for (x + a)(x + b) where ab = 12 or a + b = –7 or x(x – 4) –3(x – 4) or x(x – 3) –4(x – 3) 15(b) (5 + 3n)(x + 2y) final answer 2 B1 for 5(x + 2y) + 3n(x + 2y) or x(5 + 3n) + 2y(5 + 3n)

This question in 0580/23 Oct/Nov 2025