5 Polynomials

5 Polynomials 3. (a) m4 + 9m2n = m2(m2 + 9n)

(b) m4 + 9m2n – 9m2 – 81n
[1A]

= m2(m2 + 9n) – 9(m2 + 9n) [1M]

Conventional Questions = (m2 + 9n)(m2 – 9) [1M]
Instant Practice 1 = (m2 + 9n)(m + 3)(m – 3) [1A]
1. (a) x2 + xy – 6y2 = (x + 3y)(x – 2y) [1A]
(b) x2 + xy – 6y2 – x + 2y 4. (a) 4m4 + 8m2n + 12m2 = 4m2(m2 + 2n + 3) [1A]
= (x + 3y)(x – 2y) – (x – 2y) [1M] (b) 4m4 + 8m2n + 12m2 – m2 – 2n – 3
= (x – 2y)(x + 3y – 1) [1A] = 4m2(m2 + 2n + 3) – (m2 + 2n + 3) [1M]
= (m2 + 2n + 3)(4m2 – 1) [1M]
2. (a) 3x2 – 4xy + y2 = (3x – y)(x – y) [1A] = (m2 + 2n + 3)(2m + 1)(2m – 1) [1A]
(b) 3x2 – 4xy + y2 + 3x – y
= (3x – y)(x – y) + (3x – y) [1M] 5. (a) 8x4 + 12x3y – 4x2y = 4x2(2x2 + 3xy – y) [1A]
= (3x – y)(x – y + 1) [1A] (b) 8x4 + 12x3y – 4x2y – 18x2y2 – 27xy3 + 9y3
= 4x2(2x2 + 3xy – y) – 9y2(2x2 + 3xy – y) [1M]
3. (a) 3p2 + pq – 2q2 = (3p – 2q)(p + q) [1A] = (2x2 + 3xy – y)(4x2 – 9y2) [1M]
(b) 3p2 + pq – 2q2 – 6p – 6q = (2x2 + 3xy – y)(2x + 3y)(2x – 3y) [1A]
= (3p – 2q)(p + q) – 6(p + q) [1M]
= (p + q)(3p – 2q – 6) [1A] Instant Practice 3
1. (a)  16c2 – 24c + 9
4. (a) m – 4n = (m + 2n)(m – 2n)
2 2
[1A] = (4c – 3)2[1A]
(b) m2 – 4n2 + 2m + 4n (b) (7c – d)2 – 16c2 + 24c – 9
= (m + 2n)(m – 2n) + 2(m + 2n) [1M] = (7c – d)2 – (4c – 3)2[1M]
= (m + 2n)(m – 2n + 2) [1A] = (7c – d – 4c + 3)(7c – d + 4c – 3) [1M]
= (3c – d + 3)(11c – d – 3)[1A]
5. (a) p2 – 6pq + 9q2 = (p – 3q)2 [1A]
(b) p2 – 6pq + 9q2 – 4p + 12q 2. (a) 4a2 – 20a + 25
= (p – 3q)2 – 4(p – 3q) [1M] = (2a – 5)2[1A]
= (p – 3q)(p – 3q – 4) [1A] (b)  (3a + b)2 – 4a2 + 20b – 25
= (3a + b)2 – (2a – 5)2[1M]
6. (a) 4m2 – 12mn + 9n2 = (2m – 3n)2 [1A] = (3a + b – 2a + 5)(3a + b + 2a – 5) [1M]
(b) 4m2 – 12mn + 9n2 + 10m – 15n = (a + b + 5)(5a + b – 5)  [1A]
= (2m – 3n)2 + 5(2m – 3n) [1M]
= (2m – 3n)(2m – 3n + 5) [1A] 3. (a)  49x2 – 14x + 1
= (7x – 1)2[1A]
Instant Practice 2 (b) (9x – y)2 – 49x2 + 14x – 1
(a) x – 4x = x (x – 4) = (9x – y)2 – (7x – 1)2  [1M]
3 2 2
1. [1A]
(b) x3 – 4x2 – x + 4 = (9x – y – 7x + 1)(9x – y + 7x – 1) [1M]
= x2(x – 4) – (x – 4) [1M] = (2x – y + 1)(16x – y – 1)[1A]
= (x – 4)(x2 – 1) [1M]
= (x – 4)(x + 1)(x – 1) [1A] 4. (a)  9c2 – 30c + 25
= (3c – 5)2  [1A]
2. (a) a3 + 3a2b + a2 = a2(a + 3b + 1) [1A] (b) (4c + d) – 9c + 30c – 25
2 2

(b) a3 + 3a2b + a2 – 4a – 12b – 4 =

(4c + d)2 – (3c – 5)2[1M]
= a2(a + 3b + 1) – 4(a + 3b + 1) [1M] = (4c + d – 3c + 5)(4c + d + 3c – 5) [1M]
= (a + 3b + 1)(a2 – 4) [1M] = (c + d + 5)(7c + d – 5)[1A]
= (a + 3b + 1)(a + 2)(a – 2) [1A]

23
HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

5. (a) 49c2 + 56c + 16 3. (a) f(–2) = 7 [1M]

= (7c + 4)2  [1A] 2(–2) + k(–2) – 10(–2) + 15 = 7
3 2

(b)  (2c + 5d) – 49c – 56c – 16

2 2
k = –3
= (2c + 5d)2 – (7c + 4)2[1M] f 3 = 2 3 – 3 3 – 10 3 + 15
3 2
[1M]
= (2c + 5d – 7c – 4)(2c + 5d + 7c + 4) [1M] 2 2 2 2
=0
= (5d – 5c – 4)(9c + 5d + 4)[1A]
` 2x – 3 is a factor of f(x). [1A]
(b) f(x) = 0
Instant Practice 4 3 2
2x – 3x – 10x + 15 = 0
1. (a) f(1) = –33 [1M]
1 – 3(1) + k(1) – 36 = –33
3 2 (2x – 3)(x2 – 5) = 0 [1M]
3
k=5 x = or x2 = 5
2
f(4) = 4 – 3(4) + 5(4) – 36
3 2
[1M] 3
x = , 5 or − 5 [1M]
=0 2
` – 4 is a factor of f(x).
x [1A] ` f(x) = 0 has 3 distinct real roots.
(b) f(x) = 0 The claim is agreed. [1A]
x – 3x + 5x – 36 = 0
3 2

(x – 4)(x2 + x + 9) = 0 [1M] 4. (a) f(–1) = –20 [1M]

x = 4 or x2 + x + 9 = 0 4(–1)3 – 16(–1)2 + k(–1) + 9 = –20
For x2 + x + 9 = 0, k=9
1 1 1 1
3 2
∆ = 12 – 4(1)(9) [1M] f – =4 – – 16 – +9 – +9 [1M]
2 2 2 2
= –35
=0
0
` 2x + 1 is a factor of f(x). [1A]
` x + x + 9 = 0 has no real roots.
2
(b) f(x) = 0
` f(x) = 0 has only one real root. 3 2
4x – 16x + 9x + 9 = 0
The claim is disagreed. [1A]
(2x + 1)(2x2 – 9x + 9) = 0 [1M]
(2x + 1)(2x – 3)(x – 3) = 0
2. (a) f(2) = 24 [1M] 1 3
k(2) – 2 + 8(2) – 4 = 24
3 2 x = – , or 3 [1M]
2 2
k=2 ` f(x) = 0 has 3 real roots. [1A]
1 1 1 1
3 2
f =2 – +8 –4 [1M]
2 2 2 2 5. (a) f(1) = 7 [1M]
=0 13 + k(1)2 – 5(1) + 14 = 7
` 2x – 1 is a factor of f(x). [1A] k = –3
(b) f(x) = 0 f(2) = 2 – 3(2) – 5(2) + 14
3 2
[1M]
2x – x + 8x – 4 = 0
3 2
=0
(2x – 1)(x2 + 4) = 0 [1M] ` x – 2 is a factor of f(x). [1A]
1
x = or x2 + 4 = 0
2
For x2 + 4 = 0,
∆ = 02 – 4(1)(4) [1M]
= –16
0
` x + 4 = 0 has no real roots.
2

` f(x) = 0 has only one real root.

The claim is disagreed. [1A]

24
 5 Polynomials

(b) f(x) = 0 (b) f(x) = 0

x – 3x – 5x + 14 = 0
3 2 2
(3x – 2)(x – 2x + 3) = 0
(x – 2)(x2 – x – 7) = 0 [1M] 2
x = or x2 – 2x + 3 = 0
x = 2 or x2 – x – 7 = 0 3
For x2 – 2x + 3 = 0,
For x2 – x – 7 = 0,
∆ = (–2)2 – 4(1)(3) [1M]
− ( −1) ± ( −1)2 − 4 (1)( −7 ) = –8
x = [1M]
2 (1) 0
` x – 2x + 3 = 0 has no real roots.
2
1 ± 29 [1M]
=
2 The equation f(x) = 0 has 1 rational root. [1A]
1 ± 29
are not rational numbers.
2 2. (a) Let ax + b be the quotient when f(x) is divided by
` f(x) = 0 has only one rational root. 2x2 + 3x – 6, where a and b are constants.
The claim is disagreed. [1A] f(x) = (2x2 + 3x – 6)(ax + b) [1M]
f(–2) = –4 [1M]
` [2(–2) + 3(–2) – 6][a(–2) + b] = –4
2
3
6. (a) f – = 63 [1M] b – 2a = 1 ......(1)
2
3
3
3 3
2 f(2) = 40
2 – +3 – + k = 63
– 32 –
2 2 2 ` [2(2) + 3(2) – 6][a(2) + b] = 40
2

k = 15 b + 2a = 5 ......(2)
f(3) = 2(3)3 + 3(3)2 – 32(3) + 15 [1M] On solving (1) and (2), we have a = 1 and b = 3.
=0 ` The quotient = x + 3 [1A]
` x – 3 is a factor of f(x). [1A] (b) f(x) = 0
(b) f(x) = 0 2
(x + 3)(2x + 3x – 6) = 0
2x3 + 3x2 – 32x + 15 = 0 x = –3 or 2x2 + 3x – 6 = 0
(x – 3)(2x2 + 9x – 5) = 0 [1M] For 2x2 + 3x – 6 = 0,
(x – 3)(2x – 1)(x + 5) = 0 −3 ± 32 − 4 ( 2 )( −6 )
1 x= [1M]
x = 3, or –5 [1M] 2 (2)
2
` f(x) = 0 has 3 rational roots. [1A] −3 ± 57
=
4
Instant Practice 5 −3 ± 57
are not rational numbers. [1M]
1. (a) Let ax + b be the quotient when f(x) is divided by 4
x2 – 2x + 3, where a and b are constants. The equation f(x) = 0 has 1 rational root. [1A]
f(x) = (x2 – 2x + 3)(ax + b) [1M]
f(2) = 12 [1M] 3. (a) Let ax + b be the quotient when f(x) is divided by
` [2 – 2(2) + 3][a(2) + b] = 12
2
4x2 – 8x – 5, where a and b are constants.
2a + b = 4 ......(1) f(x) = (4x2 – 8x – 5)(ax + b) [1M]
f(–1) = –30 1
f = 36 [1M]
2
` [(–1) – 2(–1) + 3][a(–1) + b] = –30
2
1 1 1
2

b – a = –5 ......(2) ` 4 –8 –5 a + b = 36
2 2 2
On solving (1) and (2), we have a = 3 and b = –2. 1 9
a+b=–
` The quotient = 3x – 2 [1A] 2 2
a + 2b = –9 ......(1)
f(2) = 15
` [4(2) – 8(2) – 5][a(2) + b] = 15
2

2a + b = –3 ......(2)
On solving (1) and (2), we have a = 1 and b = –5.
` The quotient = x – 5 [1A]

25
HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

(b) f(x) = 0 6. (a) p(x) = (x2 + x + 1)(2x2 + x – 21) + cx + 17c[1M]

2
(x – 5)(4x – 8x – 5) = 0 [1M] p(1) = 0 [1M]
(x – 5)(2x – 5)(2x + 1) = 0 [1M] (1 + 1 + 1)[2(1 ) + 1 – 21] + c + 17c = 0
2 2

5 1 –54 + 18c = 0
x = 5,or –
2 2 c = 3[1A]
` The equation f(x) = 0 has 3 rational roots.
(b) p(x) = (x + x + 1)(2x + x – 21) + 3x + 51
2 2

The claim is disagreed. [1A]

= 2x4 + 3x3 – 18x2 – 17x + 30
p(–2)
4. (a) Let ax + b be the quotient when f(x) is divided by
= 2(–2)4 + 3(–2)3 – 18(–2)2 – 17(–2) + 30
3x2 – 3x – 5, where a and b are constants.
f(x) = (3x2 – 3x – 5)(ax + b) [1M] = 0
f(–2) = –13 [1M] ` x + 2 is a factor of p(x).[1A]
` [3(–2)2
– 3(–2) – 5][a(–2) + b] = –13 (c)  p(x) = 0
b – 2a = –1 ......(1) 2x + 3x – 18x – 17x + 30 = 0
4 3 2

f(–3) = –93 (x – 1)(x + 2)(2x2 + x – 15) = 0  [1M]

` [3(–3) – 3(–3) – 5][a(–3) + b] = –93
2
(x – 1)(x + 2)(x + 3)(2x – 5) 0 =
b – 3a = –3 ......(2) 5
x = 1, –2, –3 or [1M]
On solving (1) and (2), we have a = 2 and b = 3. 2
` The quotient = 2x + 3 [1A] 5
Note that 1, –2, –3 and are all rational numbers.
(b) f(x) = 0 2
(2x + 3)(3x2 – 3x – 5) = 0 [1M] The claim is agreed. [1A]

x=
–3
or
− ( −3) ± ( −3)2 − 4 ( 3)( −5 ) [1M] 7. (a) p(x) = (3x2 + 7x – 2)(x2 – 5) + 2cx – c + 9 [1M]
2 2 ( 3)
p(1) = 0[1M]
–3 3 ± 69 [3(1)2 + 7(1) – 2](12 – 5) + 2c – c + 9 = 0
x= or
2 6 –23 + c = 0
` The equation f(x) = 0 has 3 distinct real roots. c = 23
The claim is agreed. [1A] [1A]
(b) p(x) = (3x2 + 7x – 2)(x2 – 5) + 46x – 14
(a) Let ax + b be the quotient when f(x) is divided by
5. = 3x4 + 7x3 – 17x2 + 11x – 4
3x2 – x – 5, where a and b are constants. p(–4)
f(x) = (3x – x – 5)(ax + b)
2
[1M] = 3(–4)4 + 7(–4)3 – 17(–4)2 + 11(–4) – 4
f(–1) = 3 [1M] = 0
` [3(–1)2 – (–1) – 5][a(–1) + b] = 3
` x + 4 is a factor of p(x).[1A]
b – a = –3 ......(1)
(c) p(x) = 0
f(–2) = –36
3x + 7x – 17x + 11x – 4 = 0
4 3 2

` [3(–2)2 – (–2) – 5][a(–2) + b] = –36

(x + 4)(x – 1)(3x2 – 2x + 1) = 0  [1M]
b – 2a = –4 ......(2)
2± –8
On solving (1) and (2), we have a = 1 and b = –2. x = –4, 1 or  [1M]
6
` The quotient = x – 2 [1A]
2 ± –8
(b) f(x) = 0 Note that –4 and 1 are real numbers but
6
(x – 2)(3x2 – x – 5) = 0 [1M] are not real numbers.
x = 2 or 3x – x – 5 = 0
2
The claim is disagreed. [1A]
For 3x – x – 5 = 0,
2

–(–1) ± (–1)2 – 4(3)(–5)

x = [1M]
2(3)
1 ± 61
=
6
1 ± 61
are not rational numbers.
6
` f(x) = 0 has 1 rational root. [1A]
26
 5 Polynomials

Instant Practice 6 3. Let f(x) = (x2 + x + 1)(px + q) + 13x + 8, where p

(a)
1. Let f(x) = (x2 – 5x + 1)(px + q) + 7x + 1, where p
(a) and q are constants. [1M]
and q are constants. [1M]  Comparing the x3 terms and the constant terms, we
Comparing the x3 terms and the constant terms, we have
have p = 2 [1M]
p = 1 [1M] q + 8 = –1
q + 1 = 2 q = –9
q=1 f(x) = (x2 + x + 1)(2x – 9) + 13x + 8
f(x) = (x2 – 5x + 1)(x + 1) + 7x + 1 = 2x3 – 7x2 + 6x – 1
= x3 – 4x2 + 3x + 2 Thus, a = –7 and b = 6[1A]
Thus, a = –4 and b = 3[1A] (b) f(1) = 2(1)3 – 7(1)2 + 6(1) – 1 [1M]
(b) f(2) = 23 – 4(2)2 + 3(2) + 2 [1M] =0
=0 ` x – 1 is a factor of f(x).[1A]
` x – 2 is a factor of f(x).[1A] (c) f(x) = 0
(c) f(x) = 0 (x – 1)(2x – 5x + 1) = 0
2
[1M]
(x – 2)(x – 2x – 1) = 0
2
[1M] 5 ± 17
x = 1 or [1M]
8 4
x = 2 or 2 ± [1M]
2 5 ± 17
Note that 1 and are real numbers.
2± 8 4
Note that are irrational numbers.
2 ` All the roots of the equation f(x) = 0 are real.
The equation f(x) = 0 has two irrational roots. ` The claim is agreed. [1A]
The claim is agreed. [1A]
4. Let f(x) = (x2 – x + 3)(px + q) – 11x + 5, where p
(a)
2. Let f(x) = (x – 2x + 2)(px + q) – 3x – 1, where p
(a) 2
and q are constants. [1M]
and q are constants. [1M]  Comparing the x terms and the constant terms, we
3

 Comparing the x terms and the constant terms, we

3
have
have p = 1 [1M]
p = 3 [1M] 3q + 5 = 2
2q – 1 = –15 q = –1
q = –7 =
f(x) (x2 – x + 3)(x – 1) – 11x + 5
=
f(x) (x2 – 2x + 2)(3x – 7) – 3x – 1 = x3 – 2x2 – 7x + 2
= 3x3 – 13x2 + 17x – 15 Thus, a = –2 and b = –7[1A]
Thus, a = –13 and b = 17[1A] (b) f(–2) = (–2)3 – 2(–2)2 – 7(–2) + 2 [1M]
(b) f(3) = 3(3)3 – 13(3)2 + 17(3) – 15 [1M] 0 =
0 = ` x + 2 is a factor of f(x).[1A]
`x – 3 is a factor of f(x).[1A] (c) f(x) = 0
(c) f(x) = 0 (x + 2)(x – 4x + 1) = 0
2
[1M]
(x – 3)(3x – 4x + 5) = 0
2
[1M] 4 ± 12
x = –2 or [1M]
x = 3 or 3x2 – 4x + 5 = 0 2
For 3x2 – 4x + 5 = 0, 4 ± 12
Note that are irrational numbers.
Δ = (–4)2 – 4(3)(5) = –44 < 0  [1M] 2
The equation 3x2 – 4x + 5 = 0 has no real roots. Not all the roots of the equation f(x) = 0 are
The number of real roots of the equation f(x) = 0 rational.
is 1. The claim is disagreed. [1A]
The claim is disagreed. [1A]

27
HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

5. Let f(x) = (x2 – x + 1)(px + q) + 16x – 20, where p

(a) 2. (a) f(x) = (2x2 + mx + n)(x – 6) + (3x – 4) [1M]
and q are constants. [1M] 3 2
= 2x + (m – 12)x + (n – 6m + 3)x + (–6n – 4)
 Comparing the x3 terms and the constant terms, we ` 2x + kx – 10x + 38
3 2

have
= 2x3 + (m – 12)x2 + (n – 6m + 3)x + (–6n – 4)
p = 1 [1M]
Comparing the constant terms on both sides, we
q – 20 = –12 have
q=8 38 = –6n – 4 [1M]
f(x) = (x2 – x + 1)(x + 8) + 16x – 20 n = –7 [1A]
= x3 + 7x2 + 9x – 12 Comparing the coefficients of x on both sides, we
Thus, a = 7 and b = 9[1A] have
(b) f(–4) = (–4)3 + 7(–4)2 + 9(–4) – 12 [1M] –10 = –6m + n + 3
= 0 6m + 7 = 13
` x + 4 is a factor of f(x).[1A] m=1 [1A]
(c) f(x) = 0 (b) f(x) – g(x)
(x + 4)(x + 3x – 3) = 0
2
[1M] = (2x2 + x – 7)(x – 6) + (3x – 4) – (3x – 4)
–3 ± 21 = (2x2 + x – 7)(x – 6)
x = –4 or [1M]
2 f(x) – g(x) = 0
All the roots of the equation f(x) = 0 are real. ` (2x + x – 7)(x – 6) = 0
2
[1M]
The claim is agreed. [1A] x = 6 or 2x2 + x – 7 = 0
For 2x2 + x – 7 = 0,
Instant Practice 7 −1 ± 12 − 4 ( 2 )( −7 )
1. (a) f(x) = (x – 2)(x – 4x + q) + k
2
[1M] x= [1M]
2 (2)
= x – 6x + (8 + q)x + (–2q + k)
3 2

` x + px + 4x – 21 = x – 6x + (8 + q)x + (k – 2q)
3 2 3 2 −1 ± 57
=
Comparing the coefficients of x on both sides, we 4
have −1 ± 57
are not rational numbers.
4=8+q [1M] 4
q = –4 [1A] ` Not all the roots of the equation
(b) g(x) = f(x) – k f(x) – g(x) = 0 are rational.
= (x – 2)(x2 – 4x – 4) + k – k The claim is disagreed. [1A]
= (x – 2)(x2 – 4x – 4)
g(x) = 0 3. (a) f(x) = (x2 + ax + b)(x + 4) + (–11x + 4) [1M]
2
(x – 2)(x – 4x – 4) = 0 [1M] = x + (a + 4)x + (b + 4a – 11)x + (4b + 4)
3 2

x = 2 or x – 4x – 4 = 0 ` x + 2x – 22x + k
2 3 2

For x2 – 4x – 4 = 0, = x3 + (a + 4)x2 + (b + 4a – 11)x + (4b + 4)

Comparing the coefficients of x2 on both sides, we
− ( −4 ) ± ( −4 )2 − 4 (1)( −4 )
x= [1M] have
2 (1)
2=a+4 [1M]
4 ± 32 a = –2 [1A]
=
2 Comparing the coefficients of x on both sides, we
4 ± 32 have
are real numbers not equal to 2.
2 –22 = b + 4a – 11
` g(x) = 0 has 3 distinct real roots. b – 19 = –22
The claim is agreed. [1A] b = –3 [1A]

28
 5 Polynomials

(b) f(x) + g(x) (b) (i) Let g(x) = k(2x2 + ax – 3) + (bx + c), where k is
= (x2 – 2x – 3)(x + 4) + (–11x + 4) + (11x – 4) a non-zero constant.
= (x2 – 2x – 3)(x + 4) f(x) – g(x)
= (x – 3)(x + 1)(x + 4) = (2x2 + ax – 3)(x – 2) + (bx + c)
f(x) + g(x) = 0 – [k(2x2 + ax – 3) + (bx + c)]
(x – 3)(x + 1)(x + 4) = 0 [1M] = (2x2 + ax – 3)(x – 2 – k) [1M]
x = 3, –1 or –4 [1M] ` f(x) – g(x) is divisible by 2x2
+ ax – 3. [1A]
` f(x) + g(x) = 0 has 3 distinct real roots. (ii) f(x) – g(x) = 0
The claim is agreed. [1A] (2x – x – 3)(x – 2 – k) = 0
2

` (2x – 3)(x + 1)(x – 2 – k) = 0 [1M]

(a) f(x) = (x2 + ax + b)(x – 1) + (x + c) 3
4. [1M] x = , –1 or k + 2
2
= x + (a – 1)x + (b – a + 1)x + (c – b)
3 2
3
is not an integer.
x
` + 3x – x – 18 2
3 2

= x3 + (a – 1)x2 + (b – a + 1)x + (c – b) ` Not all the roots of the equation

f(x) – g(x) = 0 are integers.
Comparing the coefficients of x2, we have
The claim is agreed. [1A]
3=a–1 [1M]
a=4 [1A]
Comparing the coefficients of x, we have Instant Practice 8
–1 = b – a + 1 1.(a) Let mx + n be the required quotient, where m and
n are constants. [1M]
b – 3 = –1
h(x) = (mx + n)(x3 – 6x2 + 10x – 1) + mx + n
b=2 [1A]
Comparing the coefficients of x4 terms, m = 3.
(b) f(x) – g(x)
[1M]
= (x2 + 4x + 2)(x – 1) + (x + c)
Comparing the coefficients of x2 terms,
– [k(x2 + 4x + 2) + x + c]
–6n + 10m = 42
= (x2 + 4x + 2)(x – 1 – k)
n = –2
f(x) – g(x) = 0
The required quotient is 3x – 2. [1A]
(x2 + 4x + 2)(x – 1 – k) = 0 [1M] =
(b) h(x) 0
x = 1 + k or x + 4x + 2 = 0
2
(3x – 2)(x3 – 6x2 + 10x – 1) + 3x – 2 = 0
1 + k is a rational root of f(x) – g(x) = 0.
x(3x – 2)(x2 – 6x + 10) = 0 [1M]
For x2 + 4x + 2 = 0,
x = 0, 3x – 2 = 0 or x – 6x + 10 = 0
2
[1M]
−4 ± 4 2 − 4 (1)( 2 ) 2 6 ± –4
x= [1M] x = 0, or  [1M]
2 (1) 3 2
−4 ± 8 6 ± –4
Note that are not real numbers.
= 2
2
2
−4 ± 8 0 and are real numbers.
are not rational numbers. 3
2
The equation has 2 real roots. [1A]
` Not all the roots of the equation
f(x) – g(x) = 0 are rational.
The claim is disagreed. [1A] 2. (a)
Let mx + n be the required quotient, where m and
n are constants. [1M]
5. (a) f(x) = (2x2 + ax – 3)(x – 2) + (bx + c) [1M] h(x) = (mx + n)(x + 2x + 5x – 1) + mx + n
3 2

= 2x + (a – 4)x + (b – 2a – 3)x + (6 + c)
3 2 Comparing the coefficients of x4 terms, m = 5.
[1M]
` 2x – 5x – x + 6
3 2

Comparing the coefficients of x2 terms,

= 2x3 + (a – 4)x2 + (b – 2a – 3)x + (6 + c)
2n + 5m = 19
Comparing the coefficient of x2 on both sides, we
have n = –3
–5 = a – 4 [1M] The required quotient is 5x – 3.  [1A]
a = –1 [1A]
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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

(b) h(x) = 0 5. (a) Let mx + n be the required quotient, where m and

(5x – 3)(x3 + 2x2 + 5x – 1) + 5x – 3 = 0 n are constants. [1M]
x(5x – 3)(x2 + 2x + 5) = 0  [1M] =
h(x) (mx + n)(x – 6x + 5x – 2) + 2(mx + n)
3 2

x = 0, 5x – 3 = 0 or x + 2x + 5 = 0
2
[1M] Comparing the coefficients of x4 terms, m = 5.
–16 [1M]
3 –2 ±
x = 0, or  [1M] Comparing the coefficients of x2 terms,
5 2
–2 ± –16 –6n + 5m = 13
Note that are not rational numbers. n=2
2
3 The required quotient is 5x + 2. [1A]
0 and are rational numbers. (b)  h(x) = 0
5
The equation h(x) = 0 has 2 rational roots. [1A] (5x + 2)(x – 6x + 5x – 2) + 2(5x + 2) = 0
3 2

x(5x + 2)(x2 – 6x + 5) = 0 [1M]

3. (a)
Let mx + n be the required quotient, where m and x = 0, 5x + 2 = 0 or x2 – 6x + 5 = 0  [1M]
n are constants. [1M] = =
x 0, 5x + 2 0 or (x – 5)(x – 1) 0 =
h(x) = (mx + n)(x3 – 4x2 – 5x – 1) + mx + n –2
x = 0, , 1 or 5 [1M]
Comparing the coefficients of x4 terms, m = 2. 5
[1M] All the roots are rational numbers.
Comparing the coefficients of x2 terms, The equation h(x) = 0 has 4 rational roots. [1A]
–4n – 5m = 14
n = –6 Multiple-choice Questions
The required quotient is 2x – 6.  [1A]
Instant Practice 9
(b) h(x) = 0 1. ac + 2bc – 2ad – 4bd + ae + 2be
(2x – 6)(x3 – 4x2 – 5x – 1) + 2x – 6 = 0 = c(a + 2b) – 2d(a + 2b) + e(a + 2b)
x(2x – 6)(x2 – 4x – 5) = 0  [1M] = (a + 2b)(c – 2d + e)
= =
x 0, 2x – 6 0 or x – 4x – 5 0
2
= [1M] The answer is B.
x = 0, 2x – 6 = 0 or (x – 5)(x + 1) = 0
x = –1, 0, 3 or 5 [1M] 2. a2 – 3ab – 2ad + 6bd – ae + 3be
All the roots of the equation are real numbers. = a(a – 3b) – 2d(a – 3b) – e(a – 3b)
The equation h(x) = 0 has 4 real roots. [1A] = (a – 3b)(a – 2d – e)
The answer is D.
4. (a)
Let mx + n be the required quotient, where m and
n are constants. [1M]
3. 2ab + 3b2 – 4ac – 6bc + 6ad + 9bd
=
h(x) (mx + n)(x – 6x + x – 1) + mx + n
3 2

= b(2a + 3b) – 2c(2a + 3b) + 3d(2a + 3b)

Comparing the coefficients of x4 terms, m = 4.
[1M] = (2a + 3b)(b – 2c + 3d)
Comparing the coefficients of x2 terms, The answer is A.
–6n + m = 10
n = –1 4. 2mn – 4mp – 6mq – 3n2 + 6np + 9nq
The required quotient is 4x – 1. [1A] = 2m(n – 2p – 3q) – 3n(n – 2p – 3q)
(b) h(x) = 0 = (2m – 3n)(n – 2p – 3q)
(4x – 1)(x – 6x + x – 1) + 4x – 1 = 0
3 2 The answer is D.
x(4x – 1)(x2 – 6x + 1) = 0  [1M]
x = 0, 4x – 1 = 0 or x – 6x + 1 = 0
2
[1M] 5. 3pn + 6mp – qn – 4nr – 8mr – 2mq
1 6± 32 = 3p(n + 2m) – q(n + 2m) – 4r(n + 2m)
x = 0, or [1M] = (2m + n)(3p – q – 4r)
4 2
1 6 ± 32 The answer is D.
Note that 0, , are all real numbers.
4 2
The equation h(x) = 0 has 4 real roots. [1A]

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 5 Polynomials

Instant Practice 10 2. f(1) = 0

1. a – b + 2b – 2a
2 2
` 1 3
+ k (1)2
+ 3(1) +k=0
= (a2 – b2) + 2(b – a) 4 + 2k = 0
= (a + b)(a – b) – 2(a – b) k = –2
= (a – b)(a + b – 2) The remainder = f(–2)
The answer is D. = (–2)3 – 2(–2)2 + 3(–2) – 2
= –24
2. m2 + 4m – 16n2 + 16n The answer is A.
= (m2 – 16n2) + (4m + 16n)
= (m + 4n)(m – 4n) + 4(m + 4n) 3. f(1) = 0
` 16(1) + k(1) – 5 = 0
4
= (m + 4n)(m – 4n + 4)
The answer is B. k + 11 = 0
k = –11
3. a2 – 4b2 – 2a – 4b –1
The remainder = f
2
= (a2 – 4b2) – (2a + 4b)
= 16 –1 – 11 –1 – 5
4

= (a + 2b)(a – 2b) – 2(a + 2b) 2 2

= (a + 2b)(a – 2b – 2) 3
=
The answer is A. 2
The answer is C.

4. m2 – 3m – 9n2 + 9n
4. f(–1) = 0
= (m2 – 9n2) – (3m – 9n)
` 4(–1) 1000
– 2 k (–1) = 0
= (m + 3n)(m – 3n) – 3(m – 3n)
4 + 2k = 0
= (m – 3n)(m + 3n – 3)
k = –2
The answer is C.
The remainder = f(1)
= 4(1)1000 – 2(–2)(1)
5. 4m2 – 6m + 15n – 25n2
=8
= (4m2 – 25n2) – (6m – 15n)
The answer is C.
= (2m + 5n)(2m – 5n) – 3(2m – 5n)
= (2m – 5n)(2m + 5n – 3)
1
The answer is C. 5. f =0
2
1 3
1
`8 2 +k +3=0
Instant Practice 11 2
1. f(–2) = 0 4+k =0
2
` 2(–2) + 3(–2) + k = 0
3
k = –8
–22 + k = 0 The remainder = f(–1)
k = 22 = 8(–1)3 – 8(–1) + 3
The remainder = f(1) =3
= 2(1)3 + 3(1) + 22 The answer is B.
= 27
The answer is D. 6. f(1) = 0
` 1 + a(1) + b = 0
4 3

1+a+b=0
a = –1 – b
The remainder = f(–1)
= (–1)4 + a(–1)3 + b
= 1 – (–1 – b) + b
= 2b + 2
The answer is C.
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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) (New Syllabus Edition) Solution Guide

7. f(–1) = 0 5. Let p(x) = Q(x)(2x2 – 3x) + ax + b, where a and b are

` a(–1) + 20(–1) + b = 0
20 19 constants.
a – 20 + b = 0 ` p(x) = Q(x)(2x – 3)x + ax + b
b = 20 – a p(0) = b = 9
The remainder = f(1) 3 = 3a
p +9=0
2 2
= a(1)20 + 20(1)19 + b
= a + 20 + (20 – a) a = –6
= 40 The remainder is –6x + 9.
The answer is B. The answer is C.

Instant Practice 12
1. Let p(x) = Q(x)(x2 – 4) + ax + b, where a and b are
constants.
p(2) = 2a + b = 12 …… (1)
p(–2) = –2a + b = 0 …… (2)
On solving (1) and (2), we have a = 3 and b = 6.
The remainder is 3x + 6.
The answer is A.

2. Let p(x) = Q(x)(x2 – 9) + ax + b, where a and b are

constants.
p(3) = 3a + b = 24 …… (1)
p(–3) = –3a + b = 0 …… (2)
On solving (1) and (2), we have a = 4 and b = 12.
The remainder is 4x + 12.
The answer is A.

3. Let p(x) = Q(x)(x2 – 16) + ax + b, where a and b are

constants.
p(4) = 4a + b = –32 …… (1)
p(–4) = –4a + b = 0 …… (2)
On solving (1) and (2), we have a = –4 and b = –16.
The remainder is –4x – 16.
The answer is D.

4. Let p(x) = Q(x)(x2 – 3x + 2) + ax + b, where a and b are

constants.
` p(x) = Q(x)(x – 1)(x – 2) + ax + b
p(1) = a + b = 8 …… (1)
p(2) = 2a + b = 0 …… (2)
On solving (1) and (2), we have a = –8 and b = 16.
The remainder is –8x + 16.
The answer is C.

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