HONG KONG DIPLOMA OF SECONDARY EDUCATION
EXAMINATION
MATHEMATICS
Compulsory Part
SCHOOL-BASED ASSESSMENT
Sample Assessment Task
Polynomial
Marking Guidelines
教育局 課程發展處 數學教育組
Mathematics Education Section, Curriculum Development Institute
The Education Bureau of the HKSAR
Assessment Scale
The assessment scale for tasks on Mathematical Investigation is shown in the following table.
Level of
Performance
Very good
Good
Fair
Weak
Marks
Mathematical Knowledge and Investigation Skills
13–16
The student demonstrates a complete understanding of
the underlying mathematical knowledge and
investigation skills which are relevant to the task, and is
consistently competent and accurate in applying them in
handling the task. Typically, the student recognises
patterns, and states a conjecture which is justified by a
correct proof.
9–12
The student demonstrates a substantial understanding of
the underlying mathematical knowledge and
investigation skills which are relevant to the task, and is
generally competent and accurate in applying them in
handling the task. Typically, the student recognises
patterns, draws inductive generalisations to arrive at a
conjecture and attempts to provide a proof for the
conjecture.
5–8
The student demonstrates a basic understanding of the
underlying mathematical knowledge and investigation
skills which are relevant to the task, and is occasionally
competent and accurate in applying them in handling
the task. Typically, the student recognises patterns and
attempts to draw inductive generalisations to arrive at a
conjecture.
1–4
The student demonstrates a limited understanding of the
underlying mathematical knowledge and investigation
skills which are relevant to the task, and is rarely
competent and accurate in applying them in handling
the task. Typically, the student manages to recognise
simple patterns and organises results in a way that helps
pattern identifications.
Marks Mathematical Communication Skills
4
The student communicates ideas in
a clear, well organised and logically
true manner through coherent
written/verbal accounts, using
appropriate and correct forms of
mathematical
presentation
to
present conjectures and proofs.
3
The student is able to communicate
ideas
properly
through
written/verbal accounts, using
appropriate and correct forms of
mathematical presentation such as
algebraic
manipulations
and
geometric deductions.
2
The student is able to communicate
ideas with some appropriate forms
of mathematical presentation such
as
algebraic
formulae
and
geometric facts.
1
The
student
attempts
to
communicate
ideas
using
mathematical
symbols
and
notations,
diagrams,
tables,
calculations etc.

The full mark of a SBA task on Mathematical Investigation submitted should be scaled to 20 marks, of which 16 marks
are awarded for the mathematical knowledge and investigation skills while 4 marks are awarded for the communication
skills.

Teachers should base on the above assessment scale to design SBA tasks for assessing students with different abilities
and to develop the marking guidelines.
Polynomial_Marking Guidelines
2
Marking Guidelines
Solution
Performance
Part A
1.
(a)
Since f (1)  0 ,
therefore, the possible polynomial is f ( x)  k ( x  1) , where k is any non-zero
integer.
(b) Since f (2)  0 ,
therefore, the possible polynomial is f ( x)  k ( x  2) ,
where k is any non-zero integer.
(c)
2
Since f ( )  0 ,
3
therefore, the possible polynomial is f ( x)  k (3x  2) .
where k is any non-zero integer.
Evidence:
The polynomials proposed
Weak:
Only one correct
polynomial
Fair:
Two correct polynomials
Good:
Three correct polynomials
Very good: All polynomials are correct
2
(d) Since f (  )  0 ,
3
therefore, the possible polynomial is f ( x)  k (3x  2) .
where k is any non-zero integer.
Part B
2.
(a)
(i)
The root of f ( x)  0 is
Evidence:
1.
The polynomial proposed
2.
The factor found
2.
x 2
x2  2
x2  2  0
f ( x)  x 2  2
Therefore, the possible polynomial is f ( x)  k ( x 2  2) ,
where k is any non-zero integer.
Weak:
Attempt to find the
polynomial and factor, but
all are incorrect
Fair:
Correct polynomial
Good:
(a)
(ii)
Correct polynomial and
factor
Very good: Correct polynomial, factor
and working
( x 2  2)  0  x 2  2  x   2
 ( x  2 ) is the other factor.
OR
( x 2  2)  0  ( x  2 )( x  2 )  0  x   2
 ( x  2 ) is the other factor.
OR
f ( 2 )  k ((  2 ) 2  2)  k (2  2)  0
 ( x  2 ) is the other factor.
Polynomial_Marking Guidelines
3
Marking Guidelines
Solution
2.
(b) (i)
Performance
Evidence:
1. The proof stated
2. Roots found
3. Factors found
f (1  2)
 (1  2) 2  2(1  2)  1
 1 2 2  2  2  2 2 1
0
(ii)
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
Good:
All answers are correct but
minor mistakes are found in
the working
x2  2 x  1  0
x
2 44
2
 x  1 2
(iii) ( x  1  2 ) and ( x  1  2 ) are factors of the polynomial f (x) .
Very good: All answers and working
are correct
(c)
(i)
Another factor of the polynomial f (x) is ( x 
2 3
).
3

2  3 
2 3 
(ii)
x
 x 

3 
3 

 2 3 2 3 
2 3 2 3
 x 2  

 x  3  3
3
3


4
1
 x2  x 
3
9
Therefore, the possible polynomial is f ( x)  k (9 x 2  12x  1) ,
where k is any non-zero integer.
Evidence:
1. The factor found
2. The polynomial proposed
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
OR
Good:
2 3 2 3 4


3
3
3
2 3 2 3 43 1
Product of roots =



3
3
9
9
All answers correct but
minor mistakes are found in
the working or integral
coefficients are not given
Very good: All answers and working
are correct
Sum of roots =
Therefore, the possible polynomial is f ( x)  k (9 x 2  12x  1) ,
where k is any non-zero integer.
Polynomial_Marking Guidelines
4
Marking Guidelines
Solution
Performance
Part C
3.
Disagree
 f (1  2 )  0 and f (1  2 )  0 ,
 x  1  2  and  x  1  2  are factors of f (x) .
2
But  x  1  2  x  1  2   x 2  1  2   x 2  1  2 

Since f (1  2)  0 and f (1  2)  0 , from Part B, we know  x  1  2  ,
 x  1  2  ,  x  1  2  and  x  1  2  are factors of the polynomial.
 x 1  2  x 1  2  x  1  2  x  1  2 
  x2  2 x  1 x2  2 x  1
2
  x2  1  4 x2
 x4  6 x2  1
Evidence:
1. Factors found
2. Polynomial proposed
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
Good:
All answers are correct but
minor mistakes are found in
the working
Very good: All answers and working
are correct
Therefore, the possible polynomial is f ( x)  k ( x 4  6x 2  1) ,
where k is any non-zero integer.
Part D
4.
Since f ( 6  3  2 )  0 ,
( x  6  3  2 ) is a factor of polynomial f (x) .
Evidence:
1. Factors found
2. Polynomial proposed
In order to have all the coefficients of f (x) to be integers,
the following must be the factors of f (x) .
(x  6  3  2 ) , (x  6  3  2 ) , (x  6  3  2 ) ,
(x  6  3  2 ) , (x  6  3  2 ) , (x  6  3  2 ) ,
(x  6  3  2 ) , (x  6  3  2 )
The possible polynomial f (x)
= ( x  6  3  2)( x  6  3  2)( x  6  3  2)( x  6  3  2) 
( x  6  3  2)( x  6  3  2)( x  6  3  2)( x  6  3  2)
= [( x  6  3 ) 2  2][( x  6  3 ) 2  2][( x  6  3 ) 2  2][( x  6  3 ) 2  2]
= [ x2  2( 6  3) x  7  2 18][ x2  2( 6  3) x  7  2 18]
[ x2  2( 6  3) x  7  2 18][ x2  2( 6  3) x  7  2 18]
= [( x 2  7  2 18 ) 2  4( 6  3 ) 2 x 2 ][( x 2  7  2 18 ) 2  4( 6  3 ) 2 x 2 ]
= [( x2  7)2  72  4 18( x2  7)  4( 6  3)2 x2 ]
[( x2  7)2  72  4 18( x2  7)  4( 6  3)2 x2 ]
= [ x 4  (22  12 2 ) x 2  (121  84 2 ][ x 4  (22  12 2 ) x 2  (121  84 2 ]
= x 8  44 x 6  438 x 4  1292 x 2  529
Therefore, the possible polynomial is f ( x)  k ( x8  44x 6  438x 4  1292x 2  529) ,
where k is any non-zero integer.
Polynomial_Marking Guidelines
5
Weak:
Attempt the question but
all answers are incorrect
Fair:
Some mistakes are found
in the working and not all
answers are correct
Good:
All answers are correct but
minor mistakes are found in
the working
Very good: All answers and working
are correct