Mathematics · May / June 2025 · Paper 21 · Question 4
YMA01
As printed in the paper
The question as text
4. The function f (x) is defined by 3 + bx 2 + 5x – 3 f (x) = ax where a and b are constants. Given that (x + 3) is a factor of f (x),
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
f
−
3
=
0
→
×−
3
3
+ ×−
3
2
+ ×−
5
3
−=
3
0
a
b
−
27
a
+
9
b
=
18
b
=
3
a
+ *
2
1
=
7
1
3
1
2
1
7
f
→
×
+ ×
+ ×
5
−
3
=
4(b)[4]
Sets
a
b
2
4
2
2
2
4
a
+
2
b
=
18
3
2
Solves
b
=
a
+ with their
+
2
=
18
=
...,
=
...
a
b
a
b
=
2,
=
8
a
b
2
3
8
2
5
3
(
)
x
+
x
+
x
−
4(c)[3]
e.g.
2
=
...
x
+
...
x
+
...
+
Remainder
x
−
2
2
(Quotient =)
2
x
+
12
x
+
29
Remainder = 55
(a)
M1: Sets (
)
f
−
3
=
0
and proceeds to an equation in a and b Condone sign slips.
Score when you see embedded values within the equation or two correct simplified terms on the lhs of
the equation. The = 0 may be implied by further work which is not the given answer.
A1*: Completes proof with no errors including brackets and at least one intermediate stage of working such
as 27
−
+
9
=
18
and = 0 must be seen somewhere in their solution.
a
b
Alt I
x + and sets their remainder in a and b = 0. May be seen in a variety of forms
3
M1: Attempts to divide by
such as the grid or box method. Condone slips provided they are attempting to divide by
their remainder in a and b equal to 0. The = 0 may be implied by further work which is not the given
answer.
2
+
(
−
3 )
+ −
5
3(
−
3 )
ax
b
a x
b
a
+
3
3
+
2
+
5
−
3
x
ax
bx
x
3
+
3
2
ax
ax
−−
3
(15
−
9(
−
3 ))
=
0
(
b
−
3 )
a x
2
+
5
x
b
a
(
−
3 )
2
+
3(
−
3 )
b
a x
b
a x
(
)
5
−
3(
−
3 )
−
3
b
a
x
(
)
5
−
3(
−
3 )
+
15
−
9(
−
3 )
b
a
x
b
a
A1*: Completes proof with no errors seen and an intermediate stage of working such as
−
18
+
9(
−
3 )
=
0
=
3
+
2 *
(= 0 must be seen somewhere in their solution)
b
a
b
a
Alt II
M1: Substitutes in
=
3
+ , sets (
2
)
→ equation in a or b only Condone sign slips.
b
a
f
−
3
=
0
Score when you see embedded values within the equation or two correct terms on the lhs of the
equation.
A1*: Completes proof and achieves 0
=
0
concluding that
b
=
3
a
+
2
(b) Note they may have substituted in
b
=
3
a
+ early so the equation will be in a only or b only
2
1
7
=
M1: Sets
f
o.e.
and proceeds to an equation in a and b.
2
4
Condone slips provided they set their remainder in a and b equal to 7
1
−
4 o.e. (e.g.
f
2
May use algebraic division or e.g. grid method.
a
1
a
5
1
a
2
+
+
+
+
+
x
b
x
b
2
2
2
2
4
2
2
−
1
3
+
2
+
5
−
3
x
ax
bx
x
a
3
2
ax
−
x
2
a
2
5
b
+
x
+
x
5
1
a
7
−−−
3
−
+
=
2
b
2
4
2
4
a
1
a
+
2
−
+
b
x
b
x
2
2
2
1
a
5
3
+
b
+
x
−
2
2
1
5
1
a
a
5
+
+
−
−
+
b
x
b
2
2
2
4
2
or other simplified equivalent including 1
1
9
A1:
a
+
2
b
=
18
+
=
which may only be in a or only in b.
a
b
8
4
4
e.g. 7
a =
14
M1: Solves
b
=
3
a
+ with their
2
a
+
2
b
=
18
a
=
...,
b
=
...
You do not need to be concerned with the mechanics of the rearrangement. Award as long as values
appear for both a and b (both non zero) (which may be found using a calculator)
A1:
=
2,
=
8
a
b
(c)
In EPEN this is M1A1A1 but we are marking this M1A1B1
M1: Attempts to divide the cubic with their values for a and b by (
)
x −
2
and proceeds to a quadratic
quotient
ax
2
+
...
x
+
...
. where a is the value found in (b). It is not a requirement to proceed to a value for
R for this mark, but their method must allow for it to be found. If R is stated as 0 then M0.
Condone slips.
)(
)
2
3
8
2
5
3
(
2
2
..,
...,
...,
(
If they attempt
x
+
x
+
x
−=
x
−
ax
+
bx
+
c
+
R
a
=
b
=
c
=
R
=
2
2
12
29
A1: Quotient =
x
+
x
+
which can be extracted from their working e.g. on the top line of their
)(
)
algebraic division or (
2
x
−
2
2
x
+
12
x
+
29
+
R
isw
B1: Remainder = 55 which can be extracted from their working. isw
)(
)
May also be written in the expression (
2
x
−
2
2
x
+
12
x
+
29
+
55
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