BAB Pemfaktoran dan Pecahan Algebra

2 Factorisation and Algebraic Fractions

2.1 Kembangan ms.

BUKU TEKS
PBD Expansion 21 - 29

  Modul PdPc
• Kembangan ungkapan algebra ialah hasil pendaraban satu atau dua Tip Penting

.
ungkapan dalam kurungan.

hd
Expansion of algebraic expression is a product of one or two expression in brackets. a(x + y) = ax + ay
• Kembangan ungkapan algebra boleh dilakukan dengan mendarabkan
sebutan di luar kurungan dengan setiap sebutan di dalam kurungan. (a + b)(x + y)

.B
The expansion of algebraic expressions can be performed by multiplying the term outside of
the brackets with each term inside the brackets. = a(x + y) + b(x + y)
= ax + ay + bx + by

dn
1. Cari kembangan bagi setiap yang berikut berdasarkan jubin algebra yang diberikan.

S
TP2
Find the expansion of each of the following based on the given algebraic tiles.
gi
Contoh PdPc (a) 3x(2 + x) (b) (x + 5)(2x + 1)
5(x + 3)
an
2 x x 5
x 3
2x 2x2 10x
3x 6x 3x2
l

5 5x
Pe

15
1 x 5

Luas jubin
Area of tile
5(x + 3) (x + 5)(2x + 1)
3x(2 + x)
n

= 5x + 15 = 2x2 + 10x + x + 5
= 6x + 3x2
= 2x2 + 11x + 5
ta
bi

Contoh PdPc (c) (2x + 3)(x – 2) (d) (3x – 4)(2x – 3)

er

(x – 1)(x + 2) 2x 3 3x

x 2 8x –12 2x – 3
x–2 2x2 – 4x 3x – 6
n

x
2x
x–1 x2 – x 2x – 2 2 9x – 12 12 3
Pe

1 3x – 4 4

(2x + 3)(x – 2)
(x – 1)(x + 2) = 2x2 – 4x + 3x – 6
(3x – 4)(2x – 3)
= x2 – x + 2x – 2 = 2x2 – x – 6
= 6x2 – (9x – 12) – (8x – 12) – 12
= x2 + x – 2 = 6x2 – 9x + 12 – 8x + 12 – 12
Tip Penting = 6x2 – 17x + 12
–(a + b) = –a – b
–(a – b) = –a + b
–(–a + b) = a – b
–(–a – b) = a + b

SP 2.1.1 Menerangkan maksud kembangan dua ungkapan algebra.

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 Matematik  Tingkatan 2  Bab 2

2. Kembangkan setiap yang berikut. TP2

Expand each of the following.

Contoh PdPc (a) 7(4 + 5a) (b) 4p(p – 3q)

5m(7 – 2n) = 28 + 35a = 4p 2 – 12pq
= 35m – 10mn

(c) –6s(r + 4) (d) –3(y – 7w + 3) (e) – 2 x(10y – 15z + 20)

= –6rs – 24s = –3y + 21w – 9 5
= –4xy + 6xz – 8x

.
hd
.B
3. Kembangkan setiap yang berikut. TP2 Tip Penting
Expand each of the following.
(a + b)(x + y) = a(x + y) + b(x + y)
= ax + ay + bx + by

dn
Contoh PdPc (a) (a – 2b)(11 – b) (b) (k – l)(k + 2l)
(x + 1)(x + 5) = a(11 – b) – 2b(11 – b) = k(k + 2l) – l(k + 2l)

S
= x(x + 5) + 1(x + 5) = 11a – ab – 22b + 2b2 = k2 + 2kl – kl – 2l2
= x2 + 5x + x + 5
gi = k2 + kl – 2l2
= x2 + 6x + 5
l an

(c) (5 + 6e)(7f – e) (d) (3 + 2y)(1 + y) (e) (6q – 1)(4q – 3)

Pe

= 5(7f – e) + 6e(7f – e) = 3(1 + y) + 2y(1 + y) = 6q(4q – 3) – 1(4q – 3)

= 35f – 5e + 42ef – 6e2 = 3 + 3y + 2y + 2y2 = 24q2 – 18q – 4q + 3
n

= 3 + 5y + 2y2 = 24q2 – 22q + 3

ta
bi

(f ) (4a + 5b)(a – 2b) (g) (–7h + 3)(h – 5) (h) (9p + 6)(7 – 4p)
er

= 4a(a – 2b) + 5b(a – 2b) = –7h(h – 5) + 3(h – 5) = 9p(7 – 4p) + 6(7 – 4p)
= 4a2 – 8ab + 5ab – 10b2 = –7h2 + 35h + 3h – 15 = 63p – 36p2 + 42 – 24p
n

= 4a2 – 3ab – 10b2 = –7h2 + 38h – 15 = 39p – 36p2 + 42

Pe

(i) (4p – q)(p + q) (j) (2r + 2s)(3r + s) (k) (12 – 3t)(3u + t)

= 4p(p + q) – q(p + q) = 2r(3r + s) + 2s(3r + s) = 12(3u + t) – 3t(3u + t)
= 4p2 + 4pq – pq – q2 = 6r2 + 2rs + 6rs + 2s2 = 36u + 12t – 9ut – 3t2
= 4p2 + 3pq – q2 = 6r2 + 8rs + 2s2

SP 2.1.2 Melaksanakan kembangan dua ungkapan algebra.

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 Matematik  Tingkatan 2  Bab 2

4. Kembangkan setiap yang berikut. TP2

Expand each of the following.

Contoh PdPc (a) (k + 6)2 (b) (5p + 2q)2

(i) (m + n)2 = k2 + 2(k)(6) + 62 = (5p)2 + 2(5p)(2q) + (2q)2
= m2 + 2(m)(n) + n2 = k2 + 12k + 36 = 25p2 + 20pq + 4q2
= m2 + 2mn + n2

(ii) (x – y)2 (c) (3 – 4n)2 (d) (7r – 3)2

= x2 – 2(x)(y) + y2 = 32 – 2(3)(4n) + (4n)2 = (7r)2 – 2(7r)(3) + 32
= x2 – 2xy + y2 = 9 – 24n + 16n2 = 49r2 – 42r + 9

.
hd
(iii) (u + w)(u – w) (e) (2a + 9)(2a – 9) (f ) (5p – 3r)(5p + 3r)
= u2 – w2 = (2a)2 – 92 = (5p)2 – (3r)2
= 4a2 – 81 = 25p2 – 9r2

.B
dn
5. Permudahkan setiap yang berikut. TP3
Simplify each of the following.

S
Contoh PdPc
(i) (4p – 1)(3p – 2) – 12p Lakukan kembangan bagi (ii) (c – 3d)2 + 2cd
gi
menghapuskan kurungan.
= 12p2 – 8p – 3p + 2 – 12p =
Perfom expansion to remove the brackets. c2 – 6cd + 9d2 + 2cd
= 12p2 – 23p + 2 = c2 – 4cd + 9d2
an
Selesaikan sebutan serupa.
Solve the like terms.
l

(a) (a + b)(a – b) – a(a – 2b) (b) (3r + s)² + s(r + 3s)

Pe

= a2 – b2 – a2 + 2ab = 9r² + 6rs + s2 + rs + 3s2

= –b2 + 2ab = 9r² + 7rs + 4s2
n
ta

(c) a + b + 8(a + b) (d) (5 – 4n)² + 2(1 – n)

bi

= a + b + 8a + 8b = 25 – 40n + 16n² + 2 – 2n
= 9a + 9b = 27 – 42n + 16n²
n er

(e) (g + 3h)2 + (3g + 4h)(4g – 3h) (f ) (3p – 2)² – p(5p – 1)

Pe

= g2 + 6gh + 9h2 + 12g2 – 9gh + 6gh – 12h2 = 9p2 – 12p + 4 – 5p2 + p

= 13g2 + 13gh – 3h2 = 4p² – 11p + 4

2
(g) –x(x + 3y) – (x + y)² (h) 10k(7p + k) – p(p – 2k)
= –x² – 3xy – (x² + 2xy + y²) 51
= – x2 – 3xy – x2 – 2xy – y² = 2k(7p + k) – p(p – 2k)
= –2x² – 5xy – y² = 14kp + 2k² – p² + 2kp
= 16kp + 2k² – p²

SP 2.1.2 Melaksanakan kembangan dua ungkapan algebra.

SP 2.1.3 Mempermudahkan ungkapan algebra yang melibatkan gabungan operasi termasuk kembangan.

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 Matematik  Tingkatan 2  Bab 2

6. Selesaikan setiap yang berikut.

Solve each of the following.

(a) Shao Kang membeli p buah buku latihan dengan (b) Seetha mempunyai sebidang tanah berbentuk
harga RM(q + 1) sebuah dan q buah buku rujukan segi empat tepat seperti yang ditunjukkan
dengan harga RM(p + 2) sebuah. Dia membayar dalam rajah di bawah. Cari luas, dalam m2, tanah
dengan harga RM5pq. Hitung baki wang yang itu. TP4
diterimanya, dalam RM. Seetha has a piece of rectangular land as shown in the
Shao Kang bought p exercises book at RM(q + 1) each and diagram below. Find the area, in m , of the land.
2

q reference books at RM(p + 2) each. He paid at the price of

RM5pq. Calculate the balance he received, in RM.   TP4
(4x + 3) m
Baki/ Balance
= 5pq – p(q + 1) – q(p + 2)
(7x + 5) m

.
= 5pq – pq – p – pq – 2q

hd
= 3pq – p – 2q Luas tanah / Area of land (m2)
= (7x + 5)(4x + 3)
\ Baki wang/ Balance

.B
= 28x2 + 21x + 20x + 15
= RM(3pq – p – 2q)
= 28x2 + 41x + 15

dn
(c) Kelab Matematik sebuah sekolah ingin menjual roti sandwic sempena Hari Kantin. Mereka memerlukan

S
30 tin sardin dan 15 buku roti. Rajah di bawah menunjukkan harga sardin dan roti di dua buah pasar
raya. TP5 KBAT Menganalisis
gi
Mathematics Club of a school wants to sell sandwhich during Canteen’s Day. They need 30 cans of sardine and 15 loaves of bread.
The diagram shows the prices of sardine and bread in two supermarkets.
an

Pasar Raya Ekstra Pasar Raya Bajet

l
Pe

SARDIN SARDIN SARDIN

SARDIN SARDIN SARDIN SARDIN

4
n

RMs RM(r + 1) RM –– (r + 1)
RM4s 5
ta

Hitung kerugian yang dialami jika mereka membeli barang di pasaraya yang menawarkan harga yang
bi

lebih mahal. Nyatakan jawapan dalam sebutan s dan r.

Calculate the loss incurred if they bought the items in the supermarket that offers more expensive price. State the answer in terms of
er

s and r.

Jumlah bayaran di Pasar Raya Ekstra/ Total payment in Pasar Raya Ekstra
n

= 30s + 15(r + 1)
Pe

= 30s + 15r + 15

Jumlah bayaran di Pasar Raya Bajet/ Total payment in Pasar Raya Bajet
= 5(4s) + 15 × 4 (r + 1)
 
30 tin/ cans
=5
5 6 tin/ cans
= 20s + 12(r + 1)
= 20s + 12r + 12
Pasar Raya Ekstra menawarkan harga yang lebih
mahal berbanding Pasar Raya Bajet.
Kerugian/ Loss = (30s + 15r + 15) – (20s + 12r + 12) Pasar Raya Ekstra offers more expensive price compared
= 30s + 15r + 15 – 20s – 12r – 12 to Pasar Raya Bajet.
= 10s + 3r + 3

SP 2.1.4 Menyelesaikan masalah yang melibatkan kembangan dua ungkapan algebra.

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 Matematik  Tingkatan 2  Bab 2

(d) Rajah di bawah menunjukkan dua buah segi (e) Rajah di bawah menunjukkan sebuah rangka
empat tepat, ABCD dan EFCG. TP5 gambar yang diperbuat daripada kayu dan
The diagram shows two rectangles, ABCD and EFCG. plastik. Gambar tersebut berukuran 8 cm × 8 cm.
The diagram shows a picture frame made up from wood and
A (9x + 4) cm B plastic. The measurement of the picture is 8 cm × 8 cm.
(5x) cm
(5y + 1) cm Kayu
E F Wood

2y cm
D G 3x cm C
Plastik
Plastic

Cari luas, dalam cm², kawasan berlorek.

Find the area, in cm², of the shaded region. Kayu

.
(x + 2) cm

hd
Wood

Luas kawasan berlorek Plastik

Plastic
Area of shaded region
= Luas ABCD – Luas EFCG Jika lebar rangka gambar adalah sama, hitung

.B
Area of ABCD – Area of EFCG luas rangka kayu yang digunakan, dalam cm2.
= (9x + 4)(5y + 1) – (3x)(2y) If the width of the picture frame is the same, calculate the area

dn
of the wooden frame used, in cm2. TP5 KBAT Mengaplikasi
= 45xy + 9x + 20y + 4 – 6xy
Luas rangka kayu yang digunakan
= (39xy + 9x + 20y + 4) cm2 Area of the wooden frame used

S
= 2 × Luas trapezium
Area of trapezium
gi
=2×  12 × (5x + 8) × (x + 2) Hukum Kalis Sekutuan
Associative Law

= 5x + 10x + 8x + 16
2
an

= (5x2 + 18x + 16) cm2

(f ) (i) Rajah di sebelah menunjukkan jubin algebra yang dibuat oleh

l

x x
Pe

Vivian bagi kembangan ungkapan (x + y)2. Adakah Vivian membuat

kembangan tersebut dengan betul? Berikan justifikasi anda. TP6 x
The diagram shows algebraic tiles made by Vivian for the expansion of expression (x + y)2.
Does Vivian expand the expression correctly? Give your justification. Modul HEBAT M17
n
ta

Hasil tambah luas jubin/ Sum of the area of tiles x x

y
= x2 + x2 + xy + xy
x x2 x2
bi

= 2x2 + 2xy

(x + y)2 = (x + y) × (x + y)
er

y xy xy
= x2 + xy + xy + y2
= x2 + 2xy + y2
n
Pe

(x + y)2 ≠ 2x2 + 2xy. Maka, kembangan yang dilakukan adalah salah.

(x + y)2 ≠ 2x2 + 2xy. Thus, the expansion is not correct.

(ii) Bantu Vivian dengan membina jubin algebra yang betul bagi kembangan (x + y)2.
Help Vivian by constructing correct algebraic tiles for the expansion of (x + y)2.

x y

x x2 xy

(x + y)2 = x2 + xy + xy + y2
xy y = x2 + 2xy + y2 2
y

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 Matematik  Tingkatan 2  Bab 2

2.2 Pemfaktoran ms.

PBD Factorisation
BUKU TEKS 27 - 33

  Modul PdPc
• Pemfaktoran ialah proses mengenal pasti faktor sebutan dan ungkapan algebra. Tip Penting
Factorisation is the process of identifying the factors of an algebraic terms and algebraic expression.
Kembangan/ Expansion
• Faktor-faktor ini apabila didarabkan akan menghasilkan ungkapan asal.
These factors when multiplied together will form the original expression. p(p + q) = p2 + pq
• Pemfaktoran merupakan proses songsangan kepada kembangan.
Pemfaktoran/ Factorisation
Factorisation is the reversed process of an expansion.

7. Isi tempat kosong bagi setiap yang berikut. TP1

.
Peta Buih

hd
Fill in the blanks of each of the following.

(a) (b)

.B
2 3
1 3 1 a

dn
Faktor Faktor
6x bagi 6x 6 3ab bagi 3ab b
Factors Factors

S
of 6x gi of 3ab

3x x ab 3a
2x 3b
an

8. Senaraikan semua faktor sepunya bagi setiap sebutan berikut. TP2 TP3
l

List all the common factors of each of the following terms.

Pe

Contoh PdPc (a) 8y, 8y2

2x, 4xy
n

8y = 1 × 8y 8y2 = 1 × 8y2
2x = 1 × 2x 4xy = 1 × 4xy 2 × 4y 2 × 4y2
ta

2 × x 4 × xy 4 × 2y 4 × 2y2
2 × 2xy 8 × y 8 × y2
bi

2x × 2y y × 8y
x × 4y 2y × 4y
er

y × 4x
Faktor sepunya/ Common factors: Faktor sepunya/ Common factors:
n

1, 2, x dan/ and 2x 1, 2, 4, 8, y, 2y, 4y dan/ and 8y

Pe

Contoh PdPc (b) 15ef, 9df, 30f 2

9d2, 12de
15ef = 1 × 3 × 5 × e × f
9d = 1 × 3 × 3 × d × d
2 9df = 1 × 3 × 3 × d × f
12de = 1 × 2 × 2 × 3 × d × e 30f 2 = 1 × 2 × 3 × 5 × f × f

Faktor sepunya/ Common factors: Faktor sepunya/ Common factors:

1, 3 dan/ and d 1, 3 dan/ and f

SP 2.2.1 Menghubungkaitkan pendaraban ungkapan algebra dengan konsep faktor dan pemfaktoran, dan seterusnya menyenaraikan faktor bagi hasil darab
ungkapan algebra tersebut.

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 Matematik  Tingkatan 2  Bab 2

9. Tentukan FTSB bagi setiap sebutan berikut. TP2

Determine the HCF of each of the following terms.

Contoh PdPc (a) 14p, 28pq (b) 5k2lm, 25kl 2m

6y, 24yz
Tip Penting
14 14p, 28pq 5 5k2lm, 25kl 2m
FTSB bagi sebutan algebra
6 6y, 24yz p p, 2pq k k2lm, 5kl 2m
boleh ditentukan dengan
y  y, 4yz kaedah pembahgian 1, 2q l klm, 5l 2m
berulang.
 1, 4z m km, 5lm
HCF of algebraic terms can be FSTB/ HCF = 14p
determined using repeated
FSTB/ HCF = 6y division method.   k, 5l
FSTB/ HCF = 5klm

.
hd
10. Faktorkan setiap ungkapan berikut. TP3
Factorise each of the following expressions.

Contoh PdPc Contoh PdPc

.B
4x + 16 (i) a2 – 1 (ii) 2w2 – 50 Tip Penting
Faktorkan
= a – 1
2 2
= 2(w – 25) a – b
2 2 2

dahulu.

dn
4 4x + 16 FSTB/ HCF = 4 = (a + b)(a – b)
= (a + 1)(a – 1)
Factorise first. = 2(w2 – 52)
  x + 4 Maka/ Thus, = 2(w – 5)(w + 5)
4(x + 4)

(a) 14m + 21m2 (d) 81x2 – 100

S (g) 16z2 – 100
gi
7 14m + 21m2 = (9x)2 – 102 = 4(4z2 – 25)
an
m 2m + 3m2 = (9x + 10)(9x – 10) = 4[(2z)2 – 52]
2 + 3m = 4(2z + 5)(2z – 5)

l

FSTB/ HCF = 7m
Pe

Maka,
Thus,
7m(2 + 3m)
n

(b) 16y2 – 64y (e) 27m2 – 75 (h) 25a2 – 36

ta

16 16y2 – 64y = 3(9m2 – 25) = (5a)2 – 62

bi

y y2 – 4y = 3[(3m)2 – 52] = (5a + 6)(5a – 6)

y–4
er

= 3(3m – 5)(3m + 5)
FSTB/ HCF = 16y
n

Maka,
Thus,
Pe

16y(y – 4)
(c) 15p2q – 21pq2 (f ) 169u2 – 225 (i) 243g3 – 48g

3 15p2q – 21pq2 = (13u)2 – 152 = 3g(81g2 – 16)

p 5p2q – 7pq2 = (13u + 15)(13u – 15) = 3g[(9g)2 – 42]
q 5pq – 7q2 = 3g(9g – 4)(9g + 4)
5p – 7q
FSTB/ HCF = 3pq
Maka,
Thus,
3pq(5p – 7q)

SP 2.2.2 Memfaktorkan ungkapan algebra dengan pelbagai kaedah.

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 Matematik  Tingkatan 2  Bab 2

11. Faktorkan setiap ungkapan berikut. TP3

Factorise each of the following expression.

Contoh PdPc Contoh PdPc

Pendaraban silang / Cross multiplication Faktor sepunya dalam empat sebutan algebra
Common factor in four algebraic terms
y2 – 8y + 15 Pendaraban faktor 15:
Factor multiplication of 15: xy – y2 – 4x + 4y Gabungkan sebutan dengan faktor
(–1) × (–15) → (–1) + (–15) = –16 sepunya dalam satu kurungan.
(–3) × (–5) → (–3) + (–5) = –8 = (xy – y2) + (–4x + 4y)  Combine the terms with common factor
= y(x – y) – 4(x – y) in one brackets.
Pendaraban y –3 –3y = (x – y)(y – 4)
silang: (×) (+)
y –5 –5y
Cross Faktor sepunya. Faktorkan.
multiplication: y2 +15 –8y Common factor. Factorise.
   

.
hd
y2 – 8y + 15 = (y – 3)(y – 5)

(a) p2 – 4p – 12 (f ) ab + ac + bd + cd

.B
= (p + 2)(p – 6) = (ab + ac) + (bd + cd)
= a(b + c) + d(b + c)
p +2 +2p

dn
(×) (+) = (b + c)(a + d)
p –6 –6p
p2 –12 –4p

(b) 6m2 – m – 2
S (g) pq – p2 + 3q – 3p
gi
= (3m – 2)(2m + 1) = (pq – p2) + (3q – 3p)
= p(q – p) + 3(q – p)
an
3m –2 –4m
(×) (+) = (q – p)(p + 3)
2m +1 +3m
6m2 –2 –m
l
Pe

(c) –6x2 – 7x + 5 (h) bm – bn + cm – cn

= (–3x – 5)(2x – 1)
n

= (bm – bn) + (cm – cn)

–3x –5 –10x = b(m – n) + c(m – n)
ta

(×) (+)
 2x –1 +3x = (m – n)(b + c)
–6x2 +5 –7x
bi
er

(d) k2 – 12k + 36 (i) wp – hp – wq + hq

= (k – 6)2 k –6 –6k
n

= (wp – hp) – (wq – hq)

(×) (+)
k –6 –6k = p(w – h) – q(w – h)
Pe

k2 +36 –12k = (w – h)(p – q)

(e) 8t2 + 29t – 12 (j) 3h2 + 12h – 2hk – 8k

= (8t – 3)(t + 4) 8t –3 –3t = (3h2 + 12h) – (2hk + 8k)
(×) (+)
 t +4 +32t = 3h(h + 4) – 2k(h + 4)
8t2 –12 29t = (h + 4)(3h – 2k)

SP 2.2.2 Memfaktorkan ungkapan algebra dengan pelbagai kaedah.

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 Matematik  Tingkatan 2  Bab 2

12. Selesaikan setiap yang berikut.

Solve each of the following.

(a) Luas sebuah padang berbentuk segi empat (b) Hasil darab umur dua orang adik-beradik ialah
sama ialah (4x2 – 12x + 9) m2. Hitung perimeter n2 – 1. Cari beza umur mereka.   TP4
padang itu, dalam m.   TP4   The product of the ages of two brothers is n2 – 1. Find the
The area of a square field is (4x2 – 12x + 9) m2. Calculate the difference in their ages.
perimeter of the field, in m.
n2 – 1 = (n + 1)(n – 1)
4x – 12x + 9
2
Beza umur
= (2x)2 – 2(2x)(3) + 32 Difference in ages
= (2x – 3)(2x – 3) = n + 1 – (n – 1)
Panjang sisi padang = (2x – 3) m = n + 1 – n + 1
Side length of field =2

.
hd
Perimeter = 4(2x – 3)
= (8x – 12) m

.B
(c) Di sebuah kedai buah-buahan, sebiji oren berharga RMy, sebiji epal berharga 10 sen kurang daripada
sebiji oren dan sebiji pear berharga 50  sen lebih daripada sebiji oren. Mr Smith membeli 10 biji oren,
8 biji epal dan 5 biji pear. Hitung jumlah wang yang dibelanjakan, dalam RM.   TP4   KBAT Mengaplikasi

dn
At a fruit shop, an orange cost RMy, an apple cost 10 sen less than an orange and a pear cost 50 sen more than an orange. Mr Smith
bought 10 oranges, 8 apples and 5 pears. Calculate the total amount that he spent, in RM.   Modul HEBAT M17

S
Buah/ Fruit Bilangan buah/ Number of fruits Harga sebiji/ Price per (RM) RM

Oren/ Orange 10 y 10y

gi
Epal/ Apple 8 y – 0.1 8(y – 0.1)
an
Pear/ Pear 5 y + 0.5 5(y + 0.5)
Jumlah bayaran = 10y + 8(y – 0.1) + 5(y + 0.5)
l

Total payment = 10y + 8y – 0.8 + 5y + 2.5

Pe

= RM(23y + 1.7)

(d) Grace ingin memasang jubin pada lantai biliknya yang berbentuk segi empat tepat. Luas lantai biliknya
n

ialah (300xy – 150x + 900y – 450) m2. TP5   KBAT Mengaplikasi

Grace wants to tile the floor of her room which is rectangular shape. The area of the floor of her room is (300xy – 150x + 900y – 450) m2.
ta

(i) Cari dimensi sekeping jubin dalam sebutan x dan y.

Find the dimension of a piece of tile in terms of x and y.
bi

(ii) Cari bilangan jubin yang diperlukan untuk memasang seluruh lantai itu.
Find the number of tiles needed to cover the whole floor.  
er

(i) 300xy – 150x + 900y – 450 (ii) 150 keping jubin / tiles
= 150(2xy – x + 6y – 3)
n

= 150[x(2y – 1) + 3(2y – 1)]

Pe

= 150(x + 3)(2y – 1)
Dimensi jubin / Dimension of a tile
= (x + 3) m × (2y – 1) m

(e) Su Lin ingin menghitung 79 × 81 tanpa menggunakan kalkulator. Terangkan bagaimana anda dapat
membantu Su Lin menghitung hasil darab itu dengan lebih mudah menggunakan (x – 1) dan (x + 1).
Su Lin was trying to solve 79 × 81 without using a calculator. Explain how you help Su Lin solve the product more easily using (x – 1)
and (x + 1). TP6   KBAT Mencipta

(80 – 1)(80 + 1) = 802 – 12

= 6 400 – 1
= 6 399

SP 2.2.3 Menyelesaikan masalah yang melibatkan pemfaktoran.

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 Matematik  Tingkatan 2  Bab 2

2.3 Ungkapan Algebra dan Hukum Operasi Asas Aritmetik ms.

PBD Algebraic Expressions and Laws of Basic Arithmetic Operations
BUKU TEKS 34 - 38

13. Permudahkan setiap yang berikut. TP3

Simplify each of the following.

Contoh PdPc (a) (h + k)(h – k) – (h2 + k2) (b) (p + q)2 + (2p + 3q)(3p – 2q)
(i) Ungkapan algebra = h2 – k2 – h2 – k2 = p2 + 2pq + q2 + 6p2 – 4pq
Algebraic expressions = –2k2   + 9pq – 6q2
3x(x + y) + 5y(x – 2y) = 7p2 + 7pq – 5q2
= 3x2 + 3xy + 5xy – 10y2
= 3x2 + 8xy – 10y2

.
hd
(ii) Penyebut yang sama
Same denominators
(c) 3p + 2q – p – 5q (d) 4m – 3n + 3m – 4n
p – 2q p – 2q 2m + 3n 2m + 3n

.B
z – 5z = z – 5z 3p + 2q – p + 5q

2y 2y 2y = = 4m – 3n + 3m + 4n
p – 2q 2m + 3n
= –4z

dn
2p + 7q = 7m + n
2y =
p – 2q 2m + 3n
= – 2z
y

S
(iii) Penyebut dengan gandaan
gi
(e) 3n 2 + 5n2 (f ) 3h – 7h
2 2

bagi penyebut lain 12m 4m 2k 10k

an
Denominators with the multiple of
= 3n 2 + 5n2× 3 = 3h × 5 – 7h
2 2
the other denominator
5 – 3 12m 4m × 3 2k × 5 10k

= 3n + 15n = 15h – 7h
2 2
4c 2c
l
Pe

12m2 10k
= 5 – 3 × 2
Samakan
penyebut.
4c 2c × 2 = 18n2 = 8h
2
Equalise the
12m 10k
= 5 – 6 = – 1
denominators

4c 4c 4h 2
n

=
5k
ta

(iv) Penyebut tanpa faktor (g) c – 3 (h) 1 + 5

sepunya 5d 4c 7z 6z
bi

Denominators without common

factor = c × 4c – 3 × 5d = 1 × 6 + 5 × 7
5d × 4c 4c × 5d 7z × 6 6z × 7
er

3 + 5
= 4c – 15d = 6 + 35
2
4u 3v
20cd 42z
n

= 3 × 3v + 5 × 4u
4u × 3v 3v × 4u = 41
Pe

42z
= 9v + 20u
12uv

(v) Penyebut dengan faktor (i) p – 3p 3 + n

(j)
sepunya 6q 10qr 2mn 6m2
Denominators with common factor
3y + 5z = p × 5r – 3p × 3 = 3 × 3m + n × n
6q × 5r 10qr × 3 2mn × 3m 6m2 × n
4x 6x
= 5pr – 9p = 9m +2 n
2

= 3y × 3 + 5z × 2
2x 4x, 6x
2, 3 30qr 6m n
4x × 3 6x × 2 GSTK/ HCF
9y + 10z = 2x × 2 × 3

=
12x = 12x

SP 2.3.1 Melaksanakan penambahan dan penolakan ungkapan algebra yang melibatkan kembangan dan pemfaktoran.

21 © Penerbitan Pelangi Sdn. Bhd.

 Matematik  Tingkatan 2  Bab 2

14. Permudahkan setiap yang berikut. TP3

Simplify each of the following.

Contoh PdPc Contoh PdPc

Pendaraban ungkapan algebra Pembahagian ungkapan algebra
Multiplication of algebraic expressions Division of algebraic expressions
2m – 2n × 4x + 4y Faktorkan sebutan sepunya. 4b ÷ a + 1
x+y 10 Factorise common term. 2b + 1 5b Tukar ÷ kepada × dan salingan
a+1 5b
ialah .
= 4b × 5b
1 1
5b a+1
= 2(m – n) × 4(x + y) Pemudahkan
2b + 1 a + 1 Convert ÷ to × and the reciprocal of
x+y
1 5 10 Simplify a+1
is 5b .
5b a+1
4(m – n) = 20b2
= (2b + 1)(a + 1)
5

.
hd
(a) 3m × m2 + m – 6 (e) 4a – 4b ÷ a – b
m2 – 9 6m2 9b 18a
   1 1 1 2

.B
= 3m × (m – 2)(m + 3) = 4(a – b) × 18a
(m – 3)(m + 3) 6m2 9b a–b
1 2m 1 1

= (m – 2) = 8a

dn

2m(m – 3) b

S
gi
p2 + p q+2 (f ) 3q – 1 ÷ q – 5
(b) ×
2pq + 4p 3p + 3 3q – 21 4q – 28
an
1    1
p(p + 1) 1 q+2 1
= ×
2p(q + 2) 3(p + 1) = 3q – 1 × 4(q – 7)
1 1 1 3(q – 7) q–5
l

1
= 1
Pe

6 = 4(3q – 1)
3(q – 5)
n
ta

1 (g) 3x + 3y ÷ 2x – y
2 2
(c) × xy + x2y
1 – x2 4z + 2 4z + 4z + 1
bi

1 1 1 1
= × xy(1 + x)
(1 – x)(1 + x) 1 = 3(x + y) × (2z + 1)(2z + 1)
2(2z + 1) (x + y)(x – y)
er

= xy 1 1

1–x = 3(2z + 1)
n

2(x – y)
Pe

my – ny x­2 – y2 (y + 3)2 3y + 9
(d) × 2 (h) 2 ÷
x+y m – mn 16 – x 8 – 2x
1
y(m – n) 1 (x + y)(x – y) (y + 3)2 8 – 2x
= × = ×
x+y 1 m(m – n) 1 42 – x2 3y + 9
1 1
y(x – y) (y + 3)(y + 3) 2(4 – x)
= = ×
m (4 + x)(4 – x) 1 3(y + 3) 1
2(y + 3)
=
3(4 + x)

SP 2.3.2 Melaksanakan pendaraban dan pembahagian ungkapan algebra yang melibatkan kembangan dan pemfaktoran.

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 Matematik  Tingkatan 2  Bab 2

15. Selesaikan gabungan operasi berikut. TP3

Solve the following combined operations.

Contoh PdPc (a) 4(p – q)2 × 6(p + q) ÷ 12(p2 – q2)

1 + 3y – 9 × y+5
y + 3 y2 + 5y (y – 3) (y + 3) 4(p – q)2 × 6(p + q)
1
y+5
1 =
= 1 + 3(y – 3) × 12(p2 – q2)
y + 3 y(y + 5)1 (y – 3) (y + 3) 1 1 2 1
1
Faktorkan. Kemudian, 4(p – q)(p – q) × 6(p + q)
= 1 + 3
permudahkan faktor =
12(p – q)(p + q)
y + 3 y(y + 3) sepunya. 3 1
1 1 1
= y + 3 Permudahkan.
Factorise. Then, simplify
the common factors.
y(y + 3)1 Simplify. = 2(p – q)
=1
Samakan penyebut.

.
Equalise the

hd
y denominators.

.B
dn
(p + q)(p – 3q) + 3q2 (c) a + 2ab + b ÷ a + b + ab
2 2
(b)
4p – 8q 3c 3

S
1 1
p2 – 3pq + pq – 3q2 + 3q2
= (a + b)(a + b) × 3 + ab
=
gi
4(p – 2q) 3c a+b
1 1
an
p2 – 2pq
=
4(p – 2q) = a + b + ab
c
l

p­­(p – 2q) 1
= = a + b + abc
Pe

4(p – 2q) 1 c

p
=
4
n
ta
bi

(d) (m2 + 2m + 1) ÷ (m2 – 1) –

2 (e) 2px + qx + 2py + qy ÷ 8p2 + 4q
er

m+1 10x + 10y x – y2

n

m2 + 2m + 1 2
= – = x(2p + q) + y(2p + q) × (x + y)(x – y)
m –1
2
m+1 10(x + y) 4(2p + q)
Pe

m2 + 2m + 1 2(m – 1) 1 1
= –
(m + 1)(m – 1) (m + 1)(m – 1) = (x + y)(2p + q) × (x + y)(x – y)
10(x + y) 4(2p + q)
1 1
m2 + 2m + 1 – 2m + 2
=
m2 – 1 = (x + y)(x – y)
40
m2 + 3
=
= x – y
2 2
m2 – 1
40

SP 2.3.3 Melaksanakan gabungan operasi ungkapan algebra yang melibatkan kembangan dan pemfaktoran.

23 © Penerbitan Pelangi Sdn. Bhd.

 Matematik  Tingkatan 2  Bab 2

Mastery PT3 Pentaksiran Sumatif

Bahagian A Bahagian B

1. Pandakan setiap yang berikut.

1. –4(3m – n) = Match each of the following.
A –17mn C –4n + 12m
B 17mn D 4n – 12m (a + b)2 (a – 1)(a + 1)

a2 – b2 (a – b)(a + b)

.
2. (r – 5t)2 + r(7r – 2t) =

hd
A 7r 2 – 10rt + 25t 2 a2 – 1 (a – b)(a – b)
B 7r 2 + 12rt + 25t 2
C 8r 2 – 12rt + 25t 2

.B
a2 – 2ab + b2 a2 + 2ab + b2
D 8r 2 – 12rt – 25t 2
[4 markah/ 4 marks]

dn
3. Faktorkan 2. (a) Isi bulatan kosong dengan faktor bagi 8x.
Factorise Fill in the empty circles with the factors of 8x.

2m2 – 14m – 36

S
1
A 2(m – 9)(m + 2) 2
8x
gi
B 2(m + 9)(m – 2)
C (2m + 4)(2m – 9)
4x 8x 4
an
D (2m – 4)(2m + 9)

2x 8
4. Ungkapkan e – 9 ÷ e + 6e + 9 dalam bentuk
2 2
l

x
Pe

2g + 1 6g + 3
termudah.
e2 – 9 e2 + 6e + 9 [2 markah/ 2 marks]
Express ÷ in the simplest form.
2g + 1 6g + 3
(b) Pada ruang jawapan, kembangkan setiap
n

(e – 9)
A yang berikut.
(e + 9)
ta

In the answer space, expand each of the following.

(e + 9) [2 markah/ 2 marks]
B
(e – 9)
bi

Jawapan/ Answer :
3(e + 3)
C (a) 7(p + 1) 7p + 7
(e – 3)
er

3(e – 3)
D (b) y(m – 2) ym – 2y
(e + 3)
n
Pe

5. Sehamparan permaidani berbentuk heksagon

sekata diletakkan di ruang tamu. Jika panjang Bahagian C
setiap sisi permaidani itu ialah (5p + 2q) cm,
hitung perimeter permaidani tersebut. 3. (a) (i) Faktorkan 16x – 4.
A regular hexagonal carpet is placed in a living room. If the Factorise 16x – 4.
length of each side of the carpet is (5p + 2q) cm, calculate the (ii) Kembangkan:
perimeter of the carpet. Expand:
A (30pq) cm (a) 4p(q – 3)
B (32pq) cm (b) –6(3 – 4p)
C (30p + 12q) cm [3 markah/ 3 marks]
D (25 + 10p) cm Jawapan/ Answer :
(i) 4(4x – 1)
(ii) (a) 4pq – 12p
(b) –18 + 24p

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 Matematik  Tingkatan 2  Bab 2

(b) Faktorkan 16(x – 2)2 – 25(y + 1)2 dengan (b) Ungkapkan 5 – 4g – 2 sebagai satu
selengkapya. 12k 24k
Factorise 16(x – 2)2 – 25(y + 1)2 completely. pecahan tunggal dalam bentuk termudah.
[3 markah/ 3 marks] 5 4g – 2
Express – as a single fraction in its simplest
12k 24k
Jawapan/ Answer : form.
[4(x – 2) + 5(y + 1)][4(x – 2) – 5(y + 1)] [3 markah/ 3 marks]
= (4x – 8 + 5y + 5)(4x – 8 – 5y – 5) Jawapan/ Answer :
= (4x + 5y – 3)(4x – 5y – 13) 5 – 4g – 2 = 10 – (4g – 2)
12k 24k 24k
= 10 – 4g + 2
(c) (i) Permudahkan. 24k
Simplify. 12 – 4g
4p(p – 12) – 64 =
24k

.
[2 markah/ 2 marks]
= 4(3 – g)

hd
Jawapan/ Answer : 24k
4p(p – 12) – 64 3 – g
=
= 4p2 – 48p – 64 6k

.B
= 4(p2 – 12p – 16)
(c) Rajah di bawah menunjukkan dua buah segi

dn
empat tepat, A dan B.
(ii) Botol A dan botol B mempunyai panjang The diagram below shows two rectangles A and B.
yang sama.
Bottle A and bottle B have the same length.

S (2y + 3) cm A
gi
1
an
3x 6 cm (4x – 1) cm
— x 18 cm
2

Cari nilai bagi x. B (6hy – 18h) cm

l

Find the value of x.

Pe

(7hx – 21h) cm
[2 markah/ 2 marks]
Jawapan/ Answer : Nyatakan hubungan panjang dan lebar
1 x + 18 = 3x + 6 antara kedua-dua segi empat tepat A dan B.  

n

2 State the relationship of length and width between those

x + 36 = 6x + 12 two rectangles, A and B. KBAT Menilai
ta

36 – 12 = 6x – x [3 markah/ 3 marks]
24 = 5x
Tip KBAT
bi

x = 4.8
Tulis perkaitan antara panjang dan lebar dalam
Matematik.
er

Write the relationship between length and width in

4. (a) Selesaikan: Mathematics.
Solve:
n

ab – 4b ÷ a2 – 16
Jawapan/ Answer :
Pe

5b 20
Hubungan panjang dan lebar ialah luas.
[4 markah/ 4 marks] The relationship of length and width is an area.
Jawapan/ Answer : Luas A = [(2y + 3)(4x – 1)] cm2
ab – 4b ÷ a2 – 16 Area of A
5b 20 Luas B = [(6hy + 18h)(7hx + 21h)] cm2
1 1 4 Area of B
= b(a – 4) × 20
5b (a – 4)(a + 4)
1 1 1

= 4
a+4

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 Matematik  Tingkatan 2  Bab 2

y
Master
KBAT
Rajah di sebelah menunjukkan sebidang tanah berbentuk segi empat tepat A 48 – 2x F (2x) m B
yang dijadikan sebuah ladang lembu. Panjang tanah tersebut ialah 56 m dan Kolam
(2x) m Pond
lebarnya ialah 48 m. Modul HEBAT M17
E
The diagram shows a rectangular land made as a cow farm. The length of the land is 56 m and the
width is 48 m. Ladang lembu
Cow farm

(a) Hitung luas, dalam m , ladang lembu itu.

2

Calculate the area, in m2, of the cow farm. 56 – 2x

.
Jalan berturap
Luas ladang lembu/ Area of the cow farm

hd
Paved road
= Luas ABCD – Luas AFE – Luas CDE
 Area of ABCD – Area of AFE – Area of CDE D C

= 56 × 48 – 1 × 2x × (48 – 2x) – 1 × 48 × (56 – 2x)

   

.B

2 2
Tip KBAT
= 2 688 – x(48 – 2x) – 24(56 – 2x)

dn
= 2 688 – 48x + 2x2 – 1 344 + 48x (a) Luas ladang lembu
= (2x2 + 1 344) m2 = luas segi empat tepat – luas
kolam – luas jalan berturap
Area of cow farm

S
= area of rectangle – area of pond –
(b) Pemilik tanah itu akan membina pagar sepanjang EC. Nyatakan panjang, area of paved road
dalam m, pagar yang diperlukan. (b) Gunakan teorem Pythagoras
gi
The owner of the land will build a fence along EC. State the length, in m, of the fence needed. untuk mencari panjang EC.
Use the Pythagoras’ theorem to
EC2 = ED2 + DC2
an
find the length of EC.
(c) Bilangan pagar yang
EC = 
(56 – 2x)2 + 482 digunakan boleh ditentukan
=  
3 136 – 224x + 4x2 + 2 304 dengan jumlah perimeter
l

dibahagi jumlah panjang

Pe

=  4x – 224x + 5 440
2
pagar dan jaraknya.
=  4(x2 – 56x + 1 360 The number of fences needed can
be determined using the total
= (2 x2 – 56x + 1 360 ) m perimeter divided by the total fence
n

and the distance.

ta

(c) Jika pemilik tanah itu ingin memagari ABCD dengan keadaan setiap panjang pagar itu ialah 6 m dan berjarak
bi

1 m antara satu sama lain. Hitung bilangan pagar yang diperlukan. 5y

er

2y
6 1
If the owner wants to fence ABCD such that the length of the fence is m and m away from each other. Calculate the number of the
5y 2y
fences needed.
n

Perimeter ABCD/ Perimeter of ABCD = 2(56) + 2(48)

Pe

= 208 m
Bilangan pagar yang diperlukan/ Number of fences needed
= 208 = 208
6 + 1 17

5y 2y  10y  
= 208 = 208 × 10y
6×2 + 1×5 17
 5y × 2 2y × 5 
= 122 6 y
17

Praktis TIMSS/PISA
KUIZ 2

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 BAB Pemfaktoran dan Pecahan Algebra
2 Factorisation and Algebraic Fractions

Mastery PT3
Bahagian A

1. –4(3m – n) = –4(3m) – 4(–n) e2 + 6e + 9

4. e – 9 ÷
2

= –12m + 4n 2g + 1 6g + 3
Jawapan / Answer : D = (e – 3)(e + 3) ÷ (e + 3)(e + 3)
2g + 1 3(2g + 1)
1 1
= (e – 3)(e + 13) × 3(2g + 1)
2g + 1 (e 1+ 3)(e + 3)

.
2. (r – 5t)2 + r(7r – 2t)
= 3(e – 3)

hd
= (r – 5t)(r – 5t) + r(7r – 2t) (e + 3)
= r2 – 5rt – 5rt + 25t2 + 7r2 – 2rt
= 8r2 – 12rt + 25t2 Jawapan / Answer : D

.B
Jawapan / Answer : C

dn
5. Perimeter / Perimeter
= 6(5p + 2q)
3. 2m2 – 14m – 36 = (30p + 12q) cm
= 2(m2 – 7m – 18)

S
= 2(m – 9)(m + 2)
gi Jawapan / Answer : C

m –9 –9m
(×) (+)
m +2 +2m
an
m2 –18 –7m
Jawapan / Answer : A
l
Pe
n
ta
bi
n er
Pe

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