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Factor Theorem
Mark Scheme 3
Level
Subject
Exam Board
Module
Topic
Sub Topic
Booklet
A Level
Mathematics (Pure)
AQA
Core 1
Algebra
Factor Theorem
Mark Scheme 3
Time Allowed:
75 minutes
Score:
/62
Percentage:
/100
Grade Boundaries:
A*
>85%
A
777.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
Page 1 of 10
Mark schemes
1
(a)
(i)
p(–2) = –8 –16 +14 + k
p(–2) = 0
–10 + k = 0
k = 10
Must have statement if k = 10 substitute
or long division or (x + 2)(x2 – 6x + 5)
M1
AG likely withhold if p(–2) = 0 not seen
A1
2
(ii)
p(x) = (x + 2)(x2 + ….. 5)
p(x) = (x + 2)(x2 – 6x + 5)
p(x) = (x + 2)(x – 1)(x – 5)
Attempt at quadratic or second linear factor
M1
(x – 1) or (x – 5) fromͻfactorͻtheorem
A1
Must be written as product
A1
3
(b)
p(3) = 27 – 36 – 21 + k
(Remainder) = k – 30 = –20
long division scores M0
M1
Condone k – 30
A1
2
Page 2 of 10
(c)
Curve thro’ 10 marked on y-axis
B1
FT their 3 roots marked on x-axis
B1
Cubic shape with a max and min
M1
Correct graph (roughly as on left) going
beyond –2 and 5
(condone max anywhere between x = –2
and 1 and min between 1 and 5)
A1
4
[11]
2
(a)
(i)
f(1) = 1 + 4 – 5
must find f(1) NOT long division
M1
f(1) = 0
(x – 1) is factor
shown = 0 plus a statement
A1
2
Page 3 of 10
(ii)
Attempt at x2 + x + 5
long division leading to x2 ± x... or
equating coefficients
M1
f(x) = (xͻ͜ͻ᷇ӿӾx2 + x + 5)
p = 1, q = 5 by inspection scores B1, B1
A1
2
(iii)
(x =)1 is real root
B1
Consider b2 – 4ac for their x2 + x + 5
not the cubic!
M1
Hence no real roots (or only real root is 1)
CSO; all values correct plus a statement
A1
3
(b)
(i)
one term correct unsimplified
M1
second term correct unsimplified
A1
all correct unsimplified
A1
3
Page 4 of 10
(ii)
correct use of limits 1 and 2;
F(2) – F(1) attempted
M1
A1
Ⱨꜟᴒԛͻꜜᴓͻͻᶤͻ
correct unsimplified
B1
shaded area
combined integral of 7x – 6 – x3 scores
M1 for limits correctly used then
A3 correct answer with all working correct
A1
4
[14]
ͻͻͻͻͻͻͻͻͻͻӾԛӿͻͻͻͻͻӾⱳӿͻͻͻͻͻͻꜝӾ᷈ӿͻⱣͻⅎͻԑͻ᷊ͻ͜ͻ᷈᷆ͻԑͻⅎ
3
Finding p(2)
M0 long division
M1
=0,
xͻ͜ͻ᷈ͻͻͻͻͻⱳᴠͻԛͻᴓԛԝ₸ꜜꜟ
Shown = 0 AND conclusion/ statement
about x͚꞊͚ᵰ͚ᶠᶣᶧᶬᶥ͚ᶟ͚ᶤᶟᶡᶲᶭᶰ
A1
2
Page 5 of 10
(ii)
Attempt at quadratic factor
or factor theorem again for 2nd factor
M1
x2 + 3x ͜ͻ᷊
or (x+4) or (x꞊ᵯῑ͚ᶮᶰᶭᶴᶣᶢ͚ᶲᶭ͚ᶠᶣ͚ᶟ͚ᶤᶟᶡᶲᶭᶰ
A1
p(x) = (x ͜ͻ᷈ӿӾx + 4)(x ͜ͻꞋӿ
A1
3
(b)
Graph through (0,8)
8 marked
B1
Ft “their factors”
3 roots marked on x-axis
B1ft
Cubic curve through their 3 points
M1
Correct including x-intercepts correct
Condone max on y-axis etc or slightly
wrong concavity at ends of graph
A1
4
[9]
Page 6 of 10
ͻͻͻͻͻͻͻͻͻͻӾԛӿͻͻͻͻͻꜝӾ᷉ӿͻⱣͻ᷈⅍ͻ͜ͻ᷿᷉ͻԑͻ˳
4
Finding p(3) and not long division
M1
p(3) = 0
x ͜ͻ᷉ͻⱳᴠͻԛͻᴓԛԝ₸ꜜꜟ
Shown = 0
plus a statement
A1
2
(b)
x(x2ͻ͜ͻ᷊x + 3) or (xͻ͜ͻ᷉ӿӾx2ͻ͜ͻx) attempt
Or p(1) = 0
x ꞊͚ᵯ͚ᶧᶱ͚ᶟ͚ᶤᶟᶡᶲᶭᶰ͚ᶟᶲᶲᶣᶫᶮᶲ
M1
p(x) = x(xͻ͜ͻ᷇ӿӾxͻ͜ͻ᷉ӿ
Condone x + 0 or x ꞊͚ᵮ͚ᶟᶱ͚ᶤᶟᶡᶲᶭᶰ
A1
2
ӾԝӿͻͻͻͻͻӾⱳӿͻͻͻͻͻͻꜝӾ᷈ӿͻⱣͻⅎͻ͜ͻ᷿᷇ͻԑͻ᷿
Must use p(2) and not long division
M1
Ӿϸᴒꞌԛⱳꜛᴑᴒꜟͻⱳᴠӿͻ͜ͻ᷈
A1
2
Page 7 of 10
(ii)
Attempt to multiply out and compare
Or long division (2 terms of quotient)
M1
coefficients: aͻⱣͻ᷈͜
x2͚꞊͚ᵰx...
A1
b Ᵽͻ᷇͜
꞊ᵯ
A1
r Ᵽͻ᷈͜
Withhold final A1 for long division unless
written as (x͚꞊͚ᵰῑӷx2͚꞊͚ᵰx͚꞊͚ᵯῑ͚꞊͚ᵰ
A1
SC B1 for rͻⱣͻ͜ͻ᷈ͻⱳᴓͻꜙ᷆ͻᴠԝꜜꜟᴒᴑ
4
[10]
5
(a)
(i)
p(3) = 27 – 45 + 21 – 3
Finding p(3)
M1
p(3) = 0
x – 3 is a factor
Shown = 0 plus a statement
Or (x – 3) (x2 – 2x + 1)
A1
2
(ii)
B is point (3,0)
Must have coordinates
B1
1
Page 8 of 10
(b)
(i)
= 3x2 – 10x + 7
One term correct
M1
All correct with NO + c etc
A1
2
(ii)
3x2 – 10x + 7 = 0
Putting their
=0
M1
(x – 1)(3x – 7) = 0
Attempt to use quad formula or factorise
m1
at M, x =
CSO factors correct etc
A1
3
(c)
= 6x – 10
ft their
B1ft
sub x = 1,
= –4
ft their
B1ft
2
Page 9 of 10
(d)
(+c)
(i)
Increase one power by 1
M1
One term correct
A1
Two other terms correct
A1
All correct (condone missing + c)
A1
4
(ii)
Realisation that limits are 0 and 1
Condone wrong way round
B1
Attempt to sub their limits into their (d)(i)
M1
=–
CSO. Must use F(1) – F(0) correctly
A1
Area =
CSO. Convincing argument
E1
4
[18]
Page 10 of 10