Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
Factor Theorem
Mark Scheme 1
Level
Subject
Exam Board
Module
Topic
Sub Topic
Booklet
A Level
Mathematics (Pure)
AQA
Core 1
Algebra
Factor Theorem
Mark Scheme 1
Time Allowed:
83 minutes
Score:
/68
Percentage:
/100
Grade Boundaries:
A*
>85%
A
777.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
Page 1 of 11
Mark schemes
1
ӾԛӿͻͻͻͻͻꜝӾ᷇͜ӿͻⱣͻӾ᷇͜ӿ3ͻ͜ͻ᷊ͻϼӾ᷇͜ӿ2ͻ͜ͻ᷉Ӿ᷇͜ӿͻԑͻ᷇ⅎ
ᶮӷ꞊ᵯῑ͚ᶟᶲᶲᶣᶫᶮᶲᶣᶢ͚͚͚͚not long division
M1
ӾⱣͻ᷇͜ͻ᷊͜ͻԑͻ᷉ͻԑͻ᷇ⅎӿͻⱣͻ᷿᷇
A1
2
(b)
(i)
p(3) = 33ͻ͜ͻ᷊ͻϼͻ᷉2ͻ͜ͻ᷉ͻϼͻ᷉ͻԑͻ᷇ⅎ
p(3) attempted
not long division
M1
ꜝӾ᷉ӿͻⱣͻ᷈⅍ͻ͜ͻ᷿᷉ͻ͜ͻ˳ͻԑͻ᷇ⅎͻⱣͻ᷆ͻɝ xͻ͜ͻ᷉ͻⱳᴠͻԛͻᴓԛԝ₸ꜜꜟ
shown = 0 plus statement
A1
2
(ii)
Quadratic factor (x2ͻ͜ͻx + c) or (x2 + bxͻ͜ͻ᷿ӿ
꞊x͚ᶭᶰ͚꞊ᵴ͚ᶲᶣᶰᶫ͚ᶠᶷ͚ᶧᶬᶱᶮᶣᶡᶲᶧᶭᶬ
or full long division by x͚꞊͚ᵱ
or comparing coefficients
or pӷ꞊ᵰῑ͚ᶟᶲᶲᶣᶫᶮᶲᶣᶢ
M1
Quadratic factor (x2ͻ͜ͻxͻ͜ͻ᷿ӿ
correct quadratic factor (or x + 2 shown to be factor
by Factor Theorem)
A1
[p(x)= ] ( xͻ͜ͻ᷉ӿӾxͻ͜ͻ᷉ӿӾx + 2)
or [p(x)=] (x͚꞊͚ᵱῑ2(x + 2)
must see product of factors
A1
3
Page 2 of 11
(c)
cubic curve with one maximum and one minimum
M1
meeting x‫ׇ‬ᶟᶶᶧᶱ͚ᶟᶲ͚꞊ᵰ͚ᶟᶬᶢ͚ᶲᶭᶳᶡᶦᶧᶬᶥ͚x-axis at 3
A1
Final A1 is dependent on previous A1 and can be withheld if curve
has very poor curvature beyond x = 3, V shape at x = 3 etc
graph as shown, going beyond x͚ᵻ͚꞊ᵰ͚ᶠᶳᶲ͚ᶡᶭᶬᶢᶭᶬᶣ
max on or to right of y -axis
A1
3
[10]
2
ӾԛӿͻͻͻͻͻᴓӾ͜ͻ᷉ӿͻⱣͻӾ͜ͻ᷉ӿ3ͻ͜ͻ᷊xͻӾ͜ͻ᷉ӿͻԑͻ᷇᷾
ᶤӷ꞊ᵱῑ͚ᶟᶲᶲᶣᶫᶮᶲᶣᶢ͚ᶬᶭᶲ͚ᶪᶭᶬᶥ͚ᶢᶧᶴᶧᶱᶧᶭᶬ
M1
ᴓӾ͜ͻ᷉ӿͻⱣͻ͜ͻ᷈⅍ͻԑͻ᷇᷈ͻԑͻ᷇᷾ͻⱣͻ᷆ͻɝ x + 3 is a factor
shown = 0 plus statement
A1
2
(ii)
Quadratic factor (x2ͻ͜ͻ᷉x + 5)
꞊ᵱx or + 5 term by inspection or full long division attempt
(f(x) =)
(x + 3)(x2ͻ͜ͻ᷉x + 5)
must see correct product
A1
2
Page 3 of 11
(b)
(i)
4x3ͻ͜ͻ᷿᷇ͻx + 60
one of these terms correct
M1
another term correct
A1
all correct ( no +c etc)
A1
3
(ii)
4x3ͻ͜ͻ᷿᷇x + 60 = 0
must see this line OE
ɝ x3ͻ͜ͻ᷊ͻx + 15 = 0
AG
B1
1
Ӿⱳⱳⱳӿͻͻͻͻͻⱪⱳᴠԝꜟⱳꞌⱳꜛԛꜛ₸ͻꜜᴓͻꜞ₹ԛᴑꜟԛ₸ⱳԝͻⱣͻӾ͜ͻ᷉ӿ2ͻ͜ͻ᷊ͻϼͻ᷾
discriminant of “ their” quadratic or correct use of
quad eqn “formula”
M1
b2ͻ͜ͻ᷊acͻⱣͻ͜ͻ᷇᷇ͻӾꜜꜟͻb2ͻ͜ͻ᷊ac < 0 )
therefore quadratic has no (real )roots
Hence only stationary point is when xͻⱣͻ᷉͜
correct discriminant evaluated correctly
(or shown to be < 0) with appropriate conclusion
plus final statement
A1
2
(iv)
12 x2ͻ͜ͻ᷿᷇
B1
Page 4 of 11
Ᵽͻ᷇᷈Ӿ᷉͜ӿ2ͻ͜ͻ᷿᷇ͻͻͻͻͻͻͻͻӾorͻͻͻͻͻͻ᷇᷈ͻϼͻ˳ͻ͜ͻ᷿᷇ͻᴒ₸ԝӿ
sub x͚ᵻ͚꞊ᵱ͚ᶧᶬᶲᶭ͚͚᷇ᶲᶦᶣᶧᶰ͚͚᷈
M1
= 92
A1
3
(v)
Minimum since
> 0 (or 92 > 0 etc)
FT appropriate conclusion from their value from
(iv) plus reason
treat parts (iv) & (v) holistically
E1
1
[14]
3
(a)ͻͻͻͻͻꜝӾ᷈͜ӿͻⱣͻӾ᷈͜ӿ3ͻԑͻӾ᷈͜ӿ2cͻԑͻӾ᷈͜ӿdͻ͜ͻ᷇᷈
ᶮӷ꞊ᵰῑ͚ᶟᶲᶲᶣᶫᶮᶲᶣᶢ͚ᶭᶰ
long division by x + 2 as far as remainder
M1
ᶱ₸Ⱳᴒⱳꜟᶲͻ͜ⅎͻԑͻ᷊cͻ᷈͜dͻ᷇᷈͜ͻⱣͻ᷇᷾᷆͜
ᶮᶳᶲᶲᶧᶬᶥ͚ᶣᶶᶮᶰᶣᶱᶱᶧᶭᶬ͚ᶤᶭᶰ͚ᶰᶣᶫᶟᶧᶬᶢᶣᶰ͚ᵻ͚꞊ᵯᵳᵮ
M1
ɝ 2cͻ͜d + 65 = 0
AG terms all on one side in any order
(check that there are no errors in working)
A1cso
(b)
p(3) = 33 + 32 c + 3d – 12
p(3) attempted
or
long division by x – 3 as far as remainder
M1
9c + 3d + 15 = 0
any correct equation with terms collected eg 3c + d͚ᵻ꞊ᵳ
A1
Page 5 of 11
(c)
2c – d + 65 =
0
3c + d + 5 = 0
ɝ5c = – 70
Elimination of c or d
M1
ɝ cⱣͻ᷊᷇͜Ԓͻd = 37
OE
value of c or d correct unsimplified
A1
both c and d correct unsimplified
A1
3
[8]
4
(a)ͻͻͻͻͻӾⱳӿͻͻͻͻͻͻͻꜝӾ᷇͜ӿͻⱣͻӾ᷇͜ӿ3ͻԑͻ᷈Ӿ᷇͜ӿ2ͻ͜ͻ᷾Ӿ᷇͜ӿͻ͜ͻ᷿
ᶮӷ꞊ᵯῑ͚ᶟᶲᶲᶣᶫᶮᶲᶣᶢ͚not long division
M1
ꜝӾ᷇͜ӿͻⱣͻ᷇͜ͻԑͻ᷈ͻԑͻ᷾ͻ͜ͻ᷿ͻⱣͻ᷆ͻɝ x + 1 is a factor
CSO; correctly shown = 0 plus statement
A1
2
(ii)
Quad factor in this form: (x2 + bx + c)
long division as far as constant term
or comparing coefficients,
or b = 1 or c͚ᵻ͚꞊ᵴ͚ᶠᶷ͚ᶧᶬᶱᶮᶣᶡᶲᶧᶭᶬ
M1
x2 + xͻ͜ͻ᷿
correct quadratic factor
A1
[p(x) =] (x +1)(x + 3)(xͻ͜ͻ᷈ӿ
must see correct product
A1
3
Page 6 of 11
ӾԜӿͻͻͻͻꜝӾ᷆ӿͻⱣͻ᷿͜ⱡͻͻͻͻꜝӾ᷇ӿͻⱣͻ͜ͻⅎ
both p(0) and p(1) attempted
and at least one value correct
M1
ɝ p(0) > p(1)
AG both values correct plus correct statement
involving p(0) and p(1)
A1
2
(c)
cubic with one max and one min
M1
͚ᶵᶧᶲᶦ͚꞊͚ᵱ‫͚׆‬꞊͚ᵯ‫͚׆‬ᵰ͚ᶫᶟᶰᶩᶣᶢ
A1
correct with minimum to right of
y꞊ᶟᶶᶧᶱ͚ᵿᶌᶂ͚ᶥᶭᶧᶬᶥ͚ᶠᶣᶷᶭᶬᶢ͚꞊ᵱ͚ᶟᶬᶢ͚ᵰ
A1
3
[10]
5
(a)
(i)
cubic curve with one max and one min
(either way up)
M1
curve touching positive x-axis (either way up)
A1
correct graph passing through O and touching
x-axis at 2
A1
3
Page 7 of 11
(ii)
x(x2 – 4x + 4) = 3
x3 – 4x2 + 4x – 3 = 0
AG (must have = 0)
B1
1
(b)
(i)
p(–1) = (–1)3 – 4(–1)2 + 4(–1) – 3
(= – 1 – 4 – 4 – 3)
p(–1) attempted (condone one slip)
or full long division to remainder
M1
= – 12
must indicate remainder = –12
if long division used
A1
2
(ii)
p(3) = 33 – 4 × 32 + 4 × 3 – 3
p(3) attempted (condone one slip)
NOT long division
M1
p(3) = 27 – 36 + 12 – 3
p(3) = 0
– 3 is factor
shown = 0 plus statement
A1
2
(iii)
Either b = –1 (coefficient of x correct)
or c =1 (constant term correct )
allow M1 for full attempt at long division or
comparing coefficients if neither b nor c is correct
M1
p(x) = (x – 3)(x2 – x + 1)
A1
2
Page 8 of 11
(c)
Discriminant of ‘their quadratic’ = (–1)
2
–4
numerical expression must be seen
M1
ⱪⱳᴠԝꜟⱳꞌⱳꜛԛꜛ₸ͻⱣͻ͜ͻ᷉ͻӾꜜꜟͻⱢͻ᷆ӿͻ
no real roots
must have correct quadratic and statement
and all working correct
A1cso
(Only real root is x =) 3
B1
3
[13]
6
(a)
p (3) = 33ͻ͜ͻ᷈ͻϼͻ᷉2 + 3 (= 27 – 18 + 3)
p(3) attempted; not long division
M1
= 12
A1
2
(b)
p( –1) = ( –1)3ͻͻͻᶭͻ᷈ͻӾ᷇͜ӿ2 + 3
p(–1) attempted; not long division
M1
correctly shown = 0 plus statement
A1cso
2
(c)
(i)
Quadratic factor (x2 – 3x + 3)
b = –3 or c = 3 by inspection
or full long division attempt
or comparing coefficients
M1
(p (x) =)
(x +1)(x2 – 3x + 3)
must see correct product
A1
2
Page 9 of 11
(ii)
Discriminant of quadratic
b2ͻ͜ͻ᷊ԛԝͻⱣͻӾ᷉͜ӿ2ͻ͜ͻ᷊ͻϼͻ᷉
‘their’ discriminant considered possibly within
quadratic equation formula
M1
A1cso
2
[8]
7
(a)
p – 3 = (–3)3 – 13(–3) – 12
must attempt p – 3 NOT long division
M1
= –27 + 39 – 12
=0
x + 3 is factor
shown = 0 plus statement
A1
2
Page 10 of 11
(b)
x+3
x2 + bx + c
Full long division, comparing coefficients
or by inspection either b = –3 or c = –4
M1
x2 – 3x – 4 obtained
or M1 A1 for either x – 4 or x + 1
clearly found using factor theorem
A1
x+3
x–4
x+1
CSO; must be seen as a product of 3
factors
NMS full marks for correct product
SC B1 for x + 3 x – 4
or (x + 3)(x + 1)( )
or (x + 3)(x + 4)(x – 1) NMS
A1
3
[9]
Page 11 of 11