Roots of polynomial equations

3(a) y=x3 B1 Correct substitution. y 4 3 −2y−1=0 y4 =(2y+1)3 =8y3 +12y2 +6y+1 M1 Obtains an equation not involving radicals. y4 −8y3−12y2 −6y−1=0 A1 α3+β3 +γ3 +δ3 =8 B1 FT 4 © UCLES 2021 Page 7 of 14 3(b) 1 + 1 + 1 + 1 = α3β3δ3 +α3β3γ3 +β3γ3δ3 +α3γ3δ3 = 6 M1 Relates 1 + 1 + 1 + 1 to coefficients. α3 β3 γ3 δ3 α3β3γ3δ3 −1 A1 FT α3 β3 γ3 δ3 −6 A1 3 3(c) α4 +β4 +γ4 +δ4 =2(α3+β3+γ3+δ3)+4 M1 Uses original equation. =20 A1 2 Question Answer Marks Guidance

3(a) y=x3 B1 Correct substitution. y 4 3 −2y−1=0 y4 =(2y+1)3 =8y3 +12y2 +6y+1 M1 Obtains an equation not involving radicals. y4 −8y3−12y2 −6y−1=0 A1 α3+β3 +γ3 +δ3 =8 B1 FT 4 © UCLES 2021 Page 7 of 14 3(b) 1 + 1 + 1 + 1 = α3β3δ3 +α3β3γ3 +β3γ3δ3 +α3γ3δ3 = 6 M1 Relates 1 + 1 + 1 + 1 to coefficients. α3 β3 γ3 δ3 α3β3γ3δ3 −1 A1 FT α3 β3 γ3 δ3 −6 A1 3 3(c) α4 +β4 +γ4 +δ4 =2(α3+β3+γ3+δ3)+4 M1 Uses original equation. =20 A1 2 Question Answer Marks Guidance

2(a) S =2 B1 1 S =S2 −2(0) M1 Uses formula for sum of squares. 2 1 =4 A1 Correct answer implies M1A1. 3 2(b)(i) S =2S −3S B1 CAO or as a single fraction. n+3 n+2 2 n 1 2(b)(ii) S =2S −3S =2(2S −3S )−3S M1 Uses their recursive formula from part (i) to find 4 3 2 1 2 2 0 2 1 S S = 7. 4  3 2 =4 A1 2 2(c) x=2− y B1 SOI 2(2− y)3 −4(2− y)2 +3=0 M1 Makes their substitution. 2y3−8y2 +8y−3=0 A1 OE but must be an equation. 3 © UCLES 2021 Page 7 of 16 2(d) 8 M1 1 1 1 α'β'+β'γ'+γ'α' 2 Uses + + = . −3 α' β' γ' α'β'γ' 2 OR Or use 2S −8S +8S −3S =0 with substitution of their values 2 1 0 −1 8 A1 FT FT from 2(c). = 3 2 © UCLES 2021 Page 8 of 16

1 b=−(α+β+γ)=−3 B1 5=32 −2(αβ+βγ+γα) M1 A1 Uses formula for sum of squares. c=2 A1 6−3(5)+2(3)+3d =0 M1 Uses original equation or formula for sum of cubes. [Equation is x3 −3x2 +2x+1=0 ] A1 d =1 6 Question Answer Marks Guidance

1 b=−(α+β+γ)=−3 B1 5=32 −2(αβ+βγ+γα) M1 A1 Uses formula for sum of squares. c=2 A1 6−3(5)+2(3)+3d =0 M1 Uses original equation or formula for sum of cubes. [Equation is x3 −3x2 +2x+1=0 ] A1 d =1 6 Question Answer Marks Guidance

4(a) yx3 x y1 3 B1 Substitutes. 2y15y2 3 60 leading to 5y2 3 62y1125y6y23 M1 Cubes to eliminate radical. 216y3216y2 53y80 A1 Alternative method for question 4(a) (2x36)3 (5x2)3 M1 8x9 72x6 125x6 216x3 2160 yx3 leading to x3  y1 B1 Substitutes. 216y3216y2 53y80 A1 3 4(b) 333 1 33 33 33  53 B1 FT Using their answer to part (a). 216 6 6 6 12 2   53  M1 6 6 6 (3 3 3)2  216 2(33 33 33) 6 6 6 161 A1 108 3 4(c) 216S 216S 53S 24 M1 Using their 9 6 3 2169 2166 533 80 S  399 133 A1 9 216 72 2 © UCLES 2022 Page 9 of 15

4(a) yx3 x y1 3 B1 Substitutes. 2y15y2 3 60 leading to 5y2 3 62y1125y6y23 M1 Cubes to eliminate radical. 216y3216y2 53y80 A1 Alternative method for question 4(a) (2x36)3 (5x2)3 M1 8x9 72x6 125x6 216x3 2160 yx3 leading to x3  y1 B1 Substitutes. 216y3216y2 53y80 A1 3 4(b) 333 1 33 33 33  53 B1 FT Using their answer to part (a). 216 6 6 6 12 2   53  M1 6 6 6 (3 3 3)2  216 2(33 33 33) 6 6 6 161 A1 108 3 4(c) 216S 216S 53S 24 M1 Using their 9 6 3 2169 2166 533 80 S  399 133 A1 9 216 72 2 © UCLES 2022 Page 9 of 15

1(a) 2+=−b B1 2 +2=0 =−2 leading to −4+=−b M1 Solves simultaneous equations, or, express b and d in terms of  and . = 1b, =−2b A1 3 3 3 b=3=−  and d =−2. 2 2=−d leading to 4 b3 =−d leading to4b3+27d =0 M1 A1 Substitutes into third equation, AG. 27 5 1(b) 3b=b2 leading to b=3 M1 A1 Uses 22 +2 =(2+)2 −2 ( 2 +2 ) or substitutes for , in terms of b. d=−4 A1 3 © UCLES 2022 Page 6 of 15

2(a) y=x−2 leading tox= y−1 2 B1 Correct substitution. y−2+3y−1+2y−1 2 +6=0 leading to 1+3y+2y 3 2 +6y2 =0 M1 Obtains an equation not involving radicals. 36y4 +32y3 +21y2 +6y+1=0 A1 1 1 1 1 B1 FT + + + =−32 =−8 2 2 2 2 36 9 4 2(b) =6 B1 SOI 1 1 1 1 222 +222 +222 +222 M1 Relates to coefficients. + + + = 2 2 2 2 2222 222 +222222 +222 =−32 A1 3 2(c) 2 M1 Uses formula for sum of squares. 1 1 1 1  1 1 1 1  + + + = + + +  4 4 4 4 2 2 2 2  −2 ( −2−2 +−2−2 +a−2−2 +−2−2 +−2−2 +−2−2) = ( −8)2 −2 (21) 9 36 − 61 A1 162 2 © UCLES 2022 Page 7 of 15

1(a) 2+=−b B1 2 +2=0 =−2 leading to −4+=−b M1 Solves simultaneous equations, or, express b and d in terms of  and . = 1b, =−2b A1 3 3 3 b=3=−  and d =−2. 2 2=−d leading to 4 b3 =−d leading to4b3+27d =0 M1 A1 Substitutes into third equation, AG. 27 5 1(b) 3b=b2 leading to b=3 M1 A1 Uses 22 +2 =(2+)2 −2 ( 2 +2 ) or substitutes for , in terms of b. d=−4 A1 3 © UCLES 2022 Page 6 of 15

3(a) y  x2 so, x y 1 2 B1 Correct substitution. y2  y2y 1 2 50 so,  2y 1 2 2   y2  y5 2 M1 Obtains an equation which eliminates radicals. y4 2y311y2 14y250 A1 2 2 2 2 2 B1FT 4 3(b) 2222 25 B1FT 1 1 1 1 222 222 222 222 M1 Relates to coefficients.     2 2 2 2 2222 14 A1 CAO 25 3 3(c)  2 M1 Uses formula for sum of squares or uses original equation. 4 4 4 4  2 2 2 2   2 22 22 22 22 22 22 22 211 18 A1 2 © UCLES 2023 Page 8 of 16

3(a) b=−(+++)=−3 B1 5=(−3)2 −2(+++++) M1 A1 Uses formula for sum of squares. c=2 A1 +++ −d M1 Uses 6= = +++  −2 −1+−1+−1+−1 = .  d =12 A1 Equation is x4 −3x3+2x2 +12x−2=0. 6 3(b) 4+4+4+4 =3(−27)−2(5)−12(3)+2(4) M1 Uses their quartic equation derived in (a). –119 A1 2 © UCLES 2023 Page 8 of 15

4(a) y=3x+1x= 1(y−1) B1 Substitutes. 3 3 2  y−1  y−1  y−1 27 +18 +6 −1=0        3   3   3  (y−1)3 +2(y−1)2 +2(y−1)−1=0 M1  y3 −3y2 +3y−1+ 2y2 −4y+2+2y−2−1=0 Expands. y3 − y2 + y−2=0 A1 AG. 3 4(b) S =12 −2(1)=−1 M1 A1 Uses formula for sum of squares, AG. 2 S =(3+1)3+(3+1)3+(3+1)3 =−1−(1)+6 M1 Uses y3 = y2 − y+2 or expands and 3 uses original equation. 4 A1 4 4(c) (3+1)(3+1)+(3+1)(3+1)+(3+1)(3+1) 1 B1 S = = −1 (3+1)(3+1)(3+1) 2 2S =S −3+S =1−3+ 1 M1 Uses 2y−2 = y−1+ y−1. −2 1 −1 2 S =−3 A1 CAO −2 4 3 © UCLES 2023 Page 9 of 15

3(a) b=−(+++)=−3 B1 5=(−3)2 −2(+++++) M1 A1 Uses formula for sum of squares. c=2 A1 +++ −d M1 Uses 6= = +++  −2 −1+−1+−1+−1 = .  d =12 A1 Equation is x4 −3x3+2x2 +12x−2=0. 6 3(b) 4+4+4+4 =3(−27)−2(5)−12(3)+2(4) M1 Uses their quartic equation derived in (a). –119 A1 2 © UCLES 2023 Page 8 of 15

1(a) 1 p B1 2 1 1(b) 222  51 M1 Factorises. 2 2 5 A1 4 2 1(c) 4z3 2pz2 5z250 B1 FT OE FT on their value of  (in terms of p). 1 1(d) 1  1  p M1 Uses 2 2 2 2 2() 3 4 p 1 A1 CAO 12 2 © Cambridge University Press & Assessment 2024 Page 6 of 16

1(a) 1 p B1 2 1 1(b) 222  51 M1 Factorises. 2 2 5 A1 4 2 1(c) 4z32pz2 5z250 B1 FT OE FT on their value of  (in terms of p). 1 1(d) 1  1  p M1 Uses 2 2 2 2 2() 3 4 p 1 A1 CAO 12 2 © Cambridge University Press & Assessment 2024 Page 6 of 16

2(a) yx2 1 B1 Uses correct substitution. May use a different letter. (y1) 3 2 2(y1)3y11 2 10y11 2 y132y11 M1 Substitutes and obtains an equation not involving radicals. y1y22 2y12 y1 y2 4y4  4y2 4y1 OR (x3 3x)2 (2x2 1)2 x6 2x4 5x2 10 y13 2y12 5(y1)10 y3y2 4y50 A1 OE 3 2(b)  2 1 2   2 1 2   2 1 2 12(4) M1 Uses 2 2 2 ()2 2(). on their answer for part 2(a). 7 A1 CAO 2 © Cambridge University Press & Assessment 2024 Page 7 of 16 2(c)  2 1 3   2 1 3   2 1 3 7453 M1 Uses their equation from part 2(a). 4 A1 CAO Alternative method for question 2(c) Uses M1  2 1 3   2 1 3 3  2 1  2 1  2 1  3  2 1  2 1  2 1  13 31435 Uses their equation from part 2(a) in a valid formula. 4 A1 CAO 2 Question Answer Marks Guidance

3(a) y= x4 B1 Uses correct substitution. y+2y 3 4 −1=016y3 =(1−y)4 M1 Substitutes and obtains an equation not involving radicals. 16y3 =1−4y+6y2 −4y3+ y4 M1 Uses binomial expansion. y4 −20y3 +6y2 −4y+1=0 A1 Must be an equation. 4 +4 +4 +4 =20 B1 5 3(b) +++=−2 B1 x5+2x4−x=05+5+5+5 =−2(20)+(−2) M1 Multiplies original equation by x and substitutes. −42 A1 3 3(c) 8+8+8+8 =202−2(6) M1 Uses formula for sum of squares. Or alternative complete method eg another substitution. 388 A1 CAO. 2 © Cambridge University Press & Assessment 2024 Page 8 of 15

3(a) 2 +2 +2 +2 =(+++)2 −2(+++++) M1 Substitutes into formula for sum of squares. 3=22−2(+++++) +++++= 1 A1 2 2 3(b) 2(++)+2(++)+2(++)+2(++) M1 A1 Factorises and substitutes. =2(2−)+2(2−)+2(2−)+2(2−) 2(3)−4=2 A1 3 3(c)(i) 6S −12S +3S +2S +24=06S −12(4)+3(3)+2(2)+24=0 M1 Sums and substitutes. 4 3 2 1 4 A1 S = 11 A1 4 6 3 3(c)(ii) 6S −12S +3S +2S +6S =0 6S −12(11)+3(4)+2(3)+12=0 M1 Multiplies equation through by x, sums 5 4 3 2 1 5 6 and substitutes. S −2S + 1S +1S +S =0 5 4 2 3 3 2 1 S −2(11)+1(4)+1(3)+2=0 5 6 2 3 S =−4 A1 5 3 2 © Cambridge University Press & Assessment 2024 Page 7 of 16

3(a) y= x4 B1 Uses correct substitution. y+2y 3 4 −1=016y3 =(1−y)4 M1 Substitutes and obtains an equation not involving radicals. 16y3 =1−4y+6y2 −4y3+ y4 M1 Uses binomial expansion. y4 −20y3 +6y2 −4y+1=0 A1 Must be an equation. 4 +4 +4 +4 =20 B1 5 3(b) +++=−2 B1 x5+2x4−x=05+5+5+5 =−2(20)+(−2) M1 Multiplies original equation by x and substitutes. −42 A1 3 3(c) 8+8+8+8 =202−2(6) M1 Uses formula for sum of squares. Or alternative complete method eg another substitution. 388 A1 CAO. 2 © Cambridge University Press & Assessment 2024 Page 8 of 15

2(a) y=x3−1 B1 Uses correct substitution. (y+1) 3 3 +2(y+1)1 3 +1=02(y+1)1 3 =−y−2 M1 Substitutes and obtains an equation not 8(y+1)=−(y+2)3 8y+8=−y3−6y2 −12y−8 involving radicals. y3 +6y2 +20y+16=0 A1 3 2(b) ( 3−1 )2 + ( 3−1 )2 + ( 3−1 )2 =62 −2(20) M1 Uses 2 +2 +2 =(++)2 −2(++). −4 A1 FT 2 2(c) ( 3−1 )3 + ( 3−1 )3 + ( 3−1 )3 =−6(−4)−20(−6)−16(3) M1 Uses their equation from (a) or a valid formula for the sum of cubes. 96 A1 2 © Cambridge University Press & Assessment 2025 Page 10 of 18

2(a) y=x3−1 B1 Uses correct substitution. (y+1) 3 3 +2(y+1)1 3 +1=02(y+1)1 3 =−y−2 M1 Substitutes and obtains an equation not 8(y+1)=−(y+2)3 8y+8=−y3−6y2 −12y−8 involving radicals. y3 +6y2 +20y+16=0 A1 3 2(b) ( 3−1 )2 + ( 3−1 )2 + ( 3−1 )2 =62 −2(20) M1 Uses 2 +2 +2 =(++)2 −2(++). −4 A1 FT 2 2(c) ( 3−1 )3 + ( 3−1 )3 + ( 3−1 )3 =−6(−4)−20(−6)−16(3) M1 Uses their equation from (a) or a valid formula for the sum of cubes. 96 A1 2 © Cambridge University Press & Assessment 2025 Page 10 of 18

3(a) 0=2 +2 +2 +2 +2(7) M1 Uses (+++)2 =2 +2 +2 +2 + 2(+++++) 2 +2 +2 +2 =−14 A1 2 3(b) S +7S +3S +88=0 M1 Uses given equation. 4 2 1 S =10 A1FT FT from (a) 4 2 © Cambridge University Press & Assessment 2025 Page 11 of 19 3(c) ( r+2)2 =r2 +22r+4 B1 Expands. 10 ( ) M1 Collects like terms and uses results from parts  ( r+2)2 + ( r+2)2 + ( r+2)2 + ( r+2)2 (a) and (b). r=1 10 A1FT FT from (a) and (b) evaluated coefficients. =( 4r2 −28r+10 ) r=1 =4 ( 10(10+1)( 2(10)+1 )) −14(10)(10+1) M1 Applies formulae from MF19. 6 +10(10) =100 A1 Evaluated. CAO 5 Question Answer Marks Guidance

4(a) 1   M1 A1 Finds determinant. 1     1  1  = − + =1−2 −2 −2 +2  1  1     1 1−2 −2 −2 +2=02 +2 +2 =1+2(1)=3 M1 A1 Sets determinant=0 and uses =1, AG. 4 4(b) 2 +2 +2 =b2 −2c3=b2 −2c *M1 A1 Uses 2 +2 +2 =(++)2 −2(++). 3+3+3 =−b(3)−c(−b)+33=−3b+cb+3 *M1 A1 Uses cubic equation or formula for 3+3+3. c=3, b=3 DM1 A1 Solves simultaneous equations. WWW 6 Question Answer Marks Guidance

4(a) y=2x+1x= 1(y−1) B1 Correct substitution used in equation. 2 1 (y−1)4 +1(y−1)3 +1(y−1)2 +1(y−1)+1=0 M1 A1 Expands. 16 8 4 2 1 ( y4 −4y3+6y2 −4y+1 ) +1 ( y3−3y2 +3y−1 ) +1 ( y2 −2y+1 ) +1(y−1)+1=0 16 8 4 2 1 y4−1y3+1 y2+1y+11=0y4−2y3+4y2+2y+11=0 A1 Simplifies, AG. 16 8 4 8 16 4 © Cambridge University Press & Assessment 2025 Page 11 of 17 4(b) S =22 −2(4) M1 Uses 2 (+++)2 =2 +2 +2 +2 +2(+++++) S =−4 A1 2 2 4(c) S −2S +4S +2S +44=0 M1 Sums over roots. 4 3 2 1 S =−76 A1 4 2 Question Answer Marks Guidance

2(a) b=−(++)=−2 B1 3=(−2)2 −2c M1 Uses formula for sum of squares. c= 1 A1 2 3 2(b) S +bS +cS +3d =0 M1 Uses original equation or formula for 3 2 1 sum of cubes. S =−3b−2c−3d =5−3d A1FT Allow FT from part (a) for their value. 3 S +bS +cS +dS =0 M1 Uses original equation. 4 3 2 1 5+(−2)(5−3d)+ (1)(3)+2d =08d−7 =0 A1FT Allow FT from part (a) for their value 2 2 and for their value of S . 3 d = 7 A1 CAO 16 5 © Cambridge University Press & Assessment 2025 Page 10 of 19

4(a) y=2x+1x= 1(y−1) B1 Correct substitution used in equation. 2 1 (y−1)4 +1(y−1)3 +1(y−1)2 +1(y−1)+1=0 M1 A1 Expands. 16 8 4 2 1 ( y4 −4y3+6y2 −4y+1 ) +1 ( y3−3y2 +3y−1 ) +1 ( y2 −2y+1 ) +1(y−1)+1=0 16 8 4 2 1 y4−1y3+1 y2+1y+11=0y4−2y3+4y2+2y+11=0 A1 Simplifies, AG. 16 8 4 8 16 4 © Cambridge University Press & Assessment 2025 Page 11 of 17 4(b) S =22 −2(4) M1 Uses 2 (+++)2 =2 +2 +2 +2 +2(+++++) S =−4 A1 2 2 4(c) S −2S +4S +2S +44=0 M1 Sums over roots. 4 3 2 1 S =−76 A1 4 2 Question Answer Marks Guidance

2(a) =−d B1 SOI 1 M1 1 ++ =−b Uses ++=−b and replaces  with . OE   1 A1 AG. + =d−b  3 © Cambridge University Press & Assessment 2025 Page 9 of 20 2(b) d M1 Uses ++=c. −d+1− =c  1 1−c A1 AG. + =  d 2 2(c)(i) 1+3 M1 Equates expressions given in (a) and (b), with correct d−3= d2 −3d−4=0 use of coefficients. d d =4 A1 d =4only for final answer. 2 2(c)(ii) 2 +2 +2 =b2 −2c M1 2 +2 +2 =(++)2 −2(++) with correct use of coefficients. 15 A1 2 © Cambridge University Press & Assessment 2025 Page 10 of 20

4(a) EITHER © Cambridge University Press & Assessment 2026 Page 13 of 23 y = x − 2  x = y − 12  5 y − 32 − 2 y − 12 + 1 = 0   y 32 = 2 y − 5  (B1 Correct substitution seen used in equation. y 3 = ( 2 y − 5 ) 2 M1 Squares to eliminate fractional powers. y 3 − 4 y 2 + 2 0 y − 2 5 = 0 A1) Must be an equation with like terms collected. OR Alternative for 4(a) ( 5 x 2 − 2 ) x = − 1 (M1 Squares to eliminate odd powers of x. ( 5x2 −2 )2 x2 =1 ( 5 y − 1 − 2 ) 2 y − 1 = 1 B1 Correct substitution used in the equation. y3 −4y2 +20y−25=0 A1) Must be an equation with positive powers and like terms collected. 3 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) © Cambridge University Press & Assessment 2026 Page 14 of 23 2 2 2 4    − + − + − = 2 2 2 2 2 2 2 0       − − + − − + − − = B1 FT Using their answer to part (a). Must be using a cubic equation with positive powers and like terms collected. −4+−4+−4 =42−2(20) M1 2 ( 4 2 4 2 4 2 ( 2 2 2 2 2 ) 2 ) 2             − + − − − + + − − = − + − + − − − + − − = − 2 4 A1 FT FT from a cubic equation answer for 4(a) 3 4(c) S − 6 = 4 S − 4 − 2 0 S − 2 + 7 5 M1 Using 6 4 4 2 0 2 2 5 0    − − − + − − = Must use S 0 = 3 FT for their cubic from 4(a) only if using this method. S − 6 = − 1 0 1 A1 CAO S =4S −20S +25S M1 Using −8 −6 −4 −2 8 4 6 2 0 4 2 5 2 0     − − − + − − − = FT for their cubic from 4(a) only if using this method. S − 8 = 1 7 6 A1 CAO 4 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) EITHER © Cambridge University Press & Assessment 2026 Page 13 of 23 y = x − 2  x = y − 12  5 y − 32 − 2 y − 12 + 1 = 0   y 32 = 2 y − 5  (B1 Correct substitution seen used in equation. y 3 = ( 2 y − 5 ) 2 M1 Squares to eliminate fractional powers. y 3 − 4 y 2 + 2 0 y − 2 5 = 0 A1) Must be an equation with like terms collected. OR Alternative for 4(a) ( 5 x 2 − 2 ) x = − 1 (M1 Squares to eliminate odd powers of x. ( 5x2 −2 )2 x2 =1 ( 5 y − 1 − 2 ) 2 y − 1 = 1 B1 Correct substitution used in the equation. y3 −4y2 +20y−25=0 A1) Must be an equation with positive powers and like terms collected. 3 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) © Cambridge University Press & Assessment 2026 Page 14 of 23 2 2 2 4    − + − + − = 2 2 2 2 2 2 2 0       − − + − − + − − = B1 FT Using their answer to part (a). Must be using a cubic equation with positive powers and like terms collected. −4+−4+−4 =42−2(20) M1 2 ( 4 2 4 2 4 2 ( 2 2 2 2 2 ) 2 ) 2             − + − − − + + − − = − + − + − − − + − − = − 2 4 A1 FT FT from a cubic equation answer for 4(a) 3 4(c) S − 6 = 4 S − 4 − 2 0 S − 2 + 7 5 M1 Using 6 4 4 2 0 2 2 5 0    − − − + − − = Must use S 0 = 3 FT for their cubic from 4(a) only if using this method. S − 6 = − 1 0 1 A1 CAO S =4S −20S +25S M1 Using −8 −6 −4 −2 8 4 6 2 0 4 2 5 2 0     − − − + − − − = FT for their cubic from 4(a) only if using this method. S − 8 = 1 7 6 A1 CAO 4 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) © Cambridge University Press & Assessment 2026 Page 9 of 20 b ( ) 5    = − + + = − B1 − 1 9 = 5 2 − 2 c M1 Uses formula for sum of squares. c = 2 2 A1 EITHER: 1 1 = 2− 2d (M1 Uses 1 1 1 .        − + − + − = + + Using correct coefficients. d = − 2 A1FT) Follow through on their c. OR: Alternative method for the last 2 marks S − 1 2 1 1 − 9 d 5 − = S 5 1 1 ( 9 + 5 ) + 2 + 5 2 ( S 2 5 0 2 ) + ( 3 − d S − 1 ) + 1 1 2 2 ( 3 ) = d 0 = 0 (M1 Uses S 2 − 5 S 1 + 2 2 S 0 + d S − 1 = 0 and S 0 = 3 d = − 2 A1FT) Follow through on their c. Available marks 5 1(b) S +bS +cS +3d =0 M1 Uses original equation or formula for sum of 3 2 1 cubes, allow one arithmetic error. Allow FT of their b, c, and d. S 3 = − b ( − 1 9 ) − c ( 5 ) − 3 ( − 2 ) = − 1 9 9 A1FT Allow FT of their b, c, and d. 2 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a) EITHER: (B1 y=x−3x= y−1 3 Correct substitution seen in the 2y− 3 4 +3y−1+y−1 3 +1=0 equation. You may see other forms including 1 4 2+3y3 + y+ y3 =0 4 1 y3 +3y3 =−y−2 y(y+3)3 =−(y+2)3 M1 Obtains an equation not involving roots or fractional powers from a correct substitution. © Cambridge University Press & Assessment 2026 Page 11 of 25 y ( y 3 + 9 y 2 + 2 7 y + 2 7 ) = − ( y 3 + 6 y 2 + 1 2 y + 8 ) M1 Expands powers of brackets. y 4 + 1 0 y 3 + 3 3 y 2 + 3 9 y + 8 = 0 A1 Collects like terms. Answer must be an equation. −3 +−3 +−3 +−3 =−10 B1FT) FT their answer equation. OR: Alternative method for 2(a) 3 x 3 + 1 = − 2 x 4 − x ( 3 x 3 + 1 ) 3 = − ( 2 x 4 + x ) 3 (M1 Cube to make sure powers of x divide by 3 2 7 x 9 + 2 7 x 6 + 9 x 3 + 1 = − 8 x 1 2 − 1 2 x 9 − 6 x 6 − x 3 M1 Expands powers of brackets. 27y−3 +27y−2 +9y−1+1=−8y−4 −12y−3−6y−2 − y−1 B1 Correct substitution seen in the equation. y4 +10y3 +33y2 +39y+8=0 A1 Collects like terms. Answer must be an equation Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(a) © Cambridge University Press & Assessment 2026 Page 12 of 25 3 3 3 3 1 0     − + − + − + − = − B1FT) FT their answer equation. 5 2(b) 1 6 1 6 1 6 1 6 1 0 2 2 ( 3 3 )     + + + = − M1 Relates 1 6 1 6 1 6 1 6     + + + to coefficients. 3 4 A1 FT FT their answer equation from 2(a) as long as it is a quartic equation. 2 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/