Mathematics · Oct / Nov 2025 · Paper 21 · Question 4

YMA01

As printed in the paper

The question as text

4. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. 3 + 13x 2 – 10x + 8 f(x) = 4x

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Show the answer and mark scheme

4(a)[4]
2
4
x
+
21
x
+
32
x
−
2 4
x
3
+
13
x
2
−
10
x
+
8
Synthetic Division
3
2
4
x
−
8
x
2
−
21
x
10
x
2
21
x
−
42
x
32
x
+
8
32
x
−
64
72
(i)
Q x
( )
=
4
x
2
+
21
x
+
32
R = 72
(ii)
Attempts (
)
(b)(i)
−
= ×−
+
×
−
×−+
f
4
4
64 13 16 10
4 8
= −
256
+
208 40 8
+
+ =
0
Hence (x +4) is a factor *
)(
)
(
2
=
3
+
2
−
+
=
+
−
+
f( )
x
4
x
13
x
10
x
8
x
4
4
x
3
x
2
(ii)
For their
2
attempts
2
,
4
x
−
3
x
+
2
''
b
−
4
ac
''
=
9
−
32
2
2
b
−
4
ac
'
<
0
so
4
x
−
3
x
+
2
has no real roots and f ( )
x =
0
has
one at
x = −
4
(
)
(
)(
)
4(b)[5]
2
4
x
+
21
x
+
32
x
−
2 4
x
3
+
13
x
2
−
10
x
+
8
Synthetic Division
3
2
4
x
−
8
x
2
−
21
x
10
x
2
21
x
−
42
x
32
x
+
8
32
x
−
64
72
(i)
Q x
( )
=
4
x
2
+
21
x
+
32
R = 72
(ii)
Attempts (
)
(b)(i)
−
= ×−
+
×
−
×−+
f
4
4
64 13 16 10
4 8
= −
256
+
208 40 8
+
+ =
0
Hence (x +4) is a factor *
)(
)
(
2
=
3
+
2
−
+
=
+
−
+
f( )
x
4
x
13
x
10
x
8
x
4
4
x
3
x
2
(ii)
For their
2
attempts
2
,
4
x
−
3
x
+
2
''
b
−
4
ac
''
=
9
−
32
2
2
b
−
4
ac
'
<
0
so
4
x
−
3
x
+
2
has no real roots and f ( )
x =
0
has
one at
x = −
4
(
)
(
)(
)
′
=
2
+
−
=
−
+
f ( )
x
12
x
26
x
10
2 3
x
1 2
x
5
5
1
−
<
x
<
2
3
(a) (i)
x by (
)
M1:
Attempts to divide f ( )
x −
2
and achieves a 3TQ quotient of the form
4
x
2
+
bx
+
c
b
≠
0,
c
≠
0
AND a remainder that is a constant. See equivalent
form using synthetic division
A1:
Achieves a Quotient or
=
2
+
+
. It is implied if they write (i)
Q x
( )
4
x
21
x
32
2
+
+
4
x
21
x
32
It cannot just be awarded from the expression within the division sum
(a) (ii)
remainder that is a constant.
Alternatively attempts to find the value of f (2) . Allow f (2)
=
72
but if it is
incorrect you must see some evidence for the award such as showing an
intermediate line of
4
×
2
3
+
13
×
2
2
−
10
×
2
+
8
or else 32
+ (or equivalent) condoning slips. See equivalent form
+
52
−
20
8
using synthetic division
A1:
States that Remainder or R = 72. It is implied if they write (ii) 72
It cannot just be awarded from sight of 72 within the division sum BUT if a
candidate gets the division sum completely correct and does not state
and R = 72 (o.e.) they should be awarded SC 1,1,1,0
Q x
( )
=
4
x
2
+
21
x
+
32
____________________________________________________________
Alt (a) via an identity BUT other acceptable methods exist such as by inspection
which is a less formal version of this.
)(
)
(
2
Sets up a correct identity
f( )
x
=
4
x
3
+
13
x
2
−
10
x
+
8
x
−
2
Ax
+
Bx
+
C
+
R
AND then
M1:
Compares terms and achieves
A =
and values for B and C
4
A1:
States that Quotient, (i) or
Q x
( )
=
4
x
2
+
21
x
+
32
M1:
Finds R via f(2) or solving 8
with their value for C.
=
R
±
2
C
A1:
States that Remainder, (ii) or R = 72
____________________________________________________________
(b)(i)
M1: Attempts to find the value of f ( 4)
−
. For this to be awarded you must see some
values. Look for, for example,
(
)
(
)
(
)
with or without the brackets
f( 4)
−
=
4
×−
4
3
+
13
×−
4
2
−
10
×−
4
+
8
f( 4)
−
= −
256
+
208
+
40
+
8
condoning slips
You must be convinced that they have attempted f ( 4)
−
and not f ( 4)
+
by (
)
Note that dividing
=
3
+
2
−
+
x +
is M0 as it doesn’t satisfy the
f( )
x
4
x
13
x
10
x
8
4
demand of the question.
Note that simply stating f( 4)
−
=
without sight of evidence (e.g. intermediate
0
values) is M0
A1*: Requires
The award of M1
Sight of correct values or correct expression, e.g. f( 4)
−
= −
256
+
208
+
40
+
(
)
(
)
(
)
(must be correct bracketing) followed by =
f( 4)
−
=
4
×−
4
3
+
13
×−
4
2
−
10
×−
4
+
8
0
and a minimal conclusion which may be QED, hence true, ✓
(b)(ii). Mark part (b) as one complete section. Work in part (i) can count in part (ii)
Although the question states hence, these marks can be awarded independently of
(b)(i) so 00111 is possible
x by (
)
M1:
Divides f ( )
x +
and achieves an acceptable 3 term quadratic factor
4
2
If division is used look for first two terms
4
x
−
3
x
+
...
)(
If they are using inspection, look for first and last terms (
2
+
+
x
4
4
x
...
M1:
Makes a valid attempt at finding the roots or number of roots of their 3term
quadratic factor resulting from factoring out or dividing by (x+4) . Look for an
attempt at the value of
2
−
or an attempt at the correct quadratic
''
b
4
ac
''
formula o.e. such as completing the square. It is dependent upon them
2
having a quadratic factor of the form
+
+
≠
≠
4
x
bx
c b
,
0,
c
0
A1*:
This requires all of the following;
a correct division (o.e.) leading to the correct quadratic factor
2
a correct ‘proof’ showing that
−
+
hasn’t any real roots e.g.
4
x
3
x
2
2
2
−
=
−
<
or attempting to solve
−
+
=
via the quadratic
''
b
4
ac
''
9
32
0
4
x
3
x
2
0
formula followed by ‘not possible’ o.e.
a minimal reason/statement showing that f ( )
x =
0
has only one real solution.
This could be a statement that the one (real) solution is
x = − or reasons that
4
the one real solution is as a result of (
)
x +
=
4
0
There can be no incorrect statements in the body of their proof. Eg. The real solution is
(
)
x +
4
(c) This is a noncalculator question/part question
M1:
Attempts to differentiate and find the roots. This requires all of
Differentiation to a 3TQ expression with at least one correct term. FYI
(
)
f ( )
′
x
=
12
x
2
+
26
x
−
10
A valid non calculator attempt at finding the two critical values of f ( )
′
x
for a valid attempt to factorise or use the quadratic formula in an attempt to
solve f ( )
′
=
. Condone attempts in which the factor of 2 disappears. E.g.
x
0
2
+
−
=
+
−
=
12
x
26
x
10
0
(2
x
5)(3
x
1)
0
dM1: Finds the inside region for their quadratic. It is dependent upon the previous M
4(c)[3]
′
=
2
+
−
=
−
+
f ( )
x
12
x
26
x
10
2 3
x
1 2
x
5
5
1
−
<
x
<
2
3
(a) (i)
x by (
)
M1:
Attempts to divide f ( )
x −
2
and achieves a 3TQ quotient of the form
4
x
2
+
bx
+
c
b
≠
0,
c
≠
0
AND a remainder that is a constant. See equivalent
form using synthetic division
A1:
Achieves a Quotient or
=
2
+
+
. It is implied if they write (i)
Q x
( )
4
x
21
x
32
2
+
+
4
x
21
x
32
It cannot just be awarded from the expression within the division sum
(a) (ii)
remainder that is a constant.
Alternatively attempts to find the value of f (2) . Allow f (2)
=
72
but if it is
incorrect you must see some evidence for the award such as showing an
intermediate line of
4
×
2
3
+
13
×
2
2
−
10
×
2
+
8
or else 32
+ (or equivalent) condoning slips. See equivalent form
+
52
−
20
8
using synthetic division
A1:
States that Remainder or R = 72. It is implied if they write (ii) 72
It cannot just be awarded from sight of 72 within the division sum BUT if a
candidate gets the division sum completely correct and does not state
and R = 72 (o.e.) they should be awarded SC 1,1,1,0
Q x
( )
=
4
x
2
+
21
x
+
32
____________________________________________________________
Alt (a) via an identity BUT other acceptable methods exist such as by inspection
which is a less formal version of this.
)(
)
(
2
Sets up a correct identity
f( )
x
=
4
x
3
+
13
x
2
−
10
x
+
8
x
−
2
Ax
+
Bx
+
C
+
R
AND then
M1:
Compares terms and achieves
A =
and values for B and C
4
A1:
States that Quotient, (i) or
Q x
( )
=
4
x
2
+
21
x
+
32
M1:
Finds R via f(2) or solving 8
with their value for C.
=
R
±
2
C
A1:
States that Remainder, (ii) or R = 72
____________________________________________________________
(b)(i)
M1: Attempts to find the value of f ( 4)
−
. For this to be awarded you must see some
values. Look for, for example,
(
)
(
)
(
)
with or without the brackets
f( 4)
−
=
4
×−
4
3
+
13
×−
4
2
−
10
×−
4
+
8
f( 4)
−
= −
256
+
208
+
40
+
8
condoning slips
You must be convinced that they have attempted f ( 4)
−
and not f ( 4)
+
by (
)
Note that dividing
=
3
+
2
−
+
x +
is M0 as it doesn’t satisfy the
f( )
x
4
x
13
x
10
x
8
4
demand of the question.
Note that simply stating f( 4)
−
=
without sight of evidence (e.g. intermediate
0
values) is M0
A1*: Requires
The award of M1
Sight of correct values or correct expression, e.g. f( 4)
−
= −
256
+
208
+
40
+
(
)
(
)
(
)
(must be correct bracketing) followed by =
f( 4)
−
=
4
×−
4
3
+
13
×−
4
2
−
10
×−
4
+
8
0
and a minimal conclusion which may be QED, hence true, ✓
(b)(ii). Mark part (b) as one complete section. Work in part (i) can count in part (ii)
Although the question states hence, these marks can be awarded independently of
(b)(i) so 00111 is possible
x by (
)
M1:
Divides f ( )
x +
and achieves an acceptable 3 term quadratic factor
4
2
If division is used look for first two terms
4
x
−
3
x
+
...
)(
If they are using inspection, look for first and last terms (
2
+
+
x
4
4
x
...
M1:
Makes a valid attempt at finding the roots or number of roots of their 3term
quadratic factor resulting from factoring out or dividing by (x+4) . Look for an
attempt at the value of
2
−
or an attempt at the correct quadratic
''
b
4
ac
''
formula o.e. such as completing the square. It is dependent upon them
2
having a quadratic factor of the form
+
+
≠
≠
4
x
bx
c b
,
0,
c
0
A1*:
This requires all of the following;
a correct division (o.e.) leading to the correct quadratic factor
2
a correct ‘proof’ showing that
−
+
hasn’t any real roots e.g.
4
x
3
x
2
2
2
−
=
−
<
or attempting to solve
−
+
=
via the quadratic
''
b
4
ac
''
9
32
0
4
x
3
x
2
0
formula followed by ‘not possible’ o.e.
a minimal reason/statement showing that f ( )
x =
0
has only one real solution.
This could be a statement that the one (real) solution is
x = − or reasons that
4
the one real solution is as a result of (
)
x +
=
4
0
There can be no incorrect statements in the body of their proof. Eg. The real solution is
(
)
x +
4
(c) This is a noncalculator question/part question
M1:
Attempts to differentiate and find the roots. This requires all of
Differentiation to a 3TQ expression with at least one correct term. FYI
(
)
f ( )
′
x
=
12
x
2
+
26
x
−
10
A valid non calculator attempt at finding the two critical values of f ( )
′
x
for a valid attempt to factorise or use the quadratic formula in an attempt to
solve f ( )
′
=
. Condone attempts in which the factor of 2 disappears. E.g.
x
0
2
+
−
=
+
−
=
12
x
26
x
10
0
(2
x
5)(3
x
1)
0
dM1: Finds the inside region for their quadratic. It is dependent upon the previous M

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