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Matematik Tingkatan 2 Bab 2
9. Tentukan FSTB bagi setiap sebutan berikut. SP 2.2.2 TP2
Determine the HCF of each of the following terms.
(a) 14p, 28pq (b) 5k lm, 25kl m
2
2
6y, 24yz
2
2
Tip Penting 14 14p, 28pq 5 5k lm, 25kl m
6 6y, 24yz FSTB bagi sebutan algebra p p, 2pq k k lm, 5kl m
2
2
y y, 4yz boleh ditentukan dengan 1, 2q l klm, 5l m
2
1, 4z kaedah pembahagian
berulang. FSTB/ HCF = 14p m km, 5lm
HCF of algebraic terms can be
FSTB/ HCF = 6y determined using repeated k, 5l
division method.
FSTB/ HCF = 5klm
10. Faktorkan setiap ungkapan berikut. SP 2.2.2 TP3
Factorise each of the following expressions.

4x + 16 (i) a – 1 (ii) 2w – 50 Tip Penting
2
2
= a – 1 Faktorkan = 2(w – 25) a – b 2
2
2
2
2
dahulu.
4 4x + 16 FSTB/ HCF = 4 = (a + 1)(a – 1) Factorise first. = 2(w – 5 ) = (a + b)(a – b)
2
2
x + 4 Maka/ Thus, = 2(w – 5)(w + 5)
4(x + 4)
(a) 14m + 21m 2 (d) 81x – 100 (g) 16z – 100
2
2
7 14m + 21m 2 = (9x) – 10 2 = 4(4z – 25)
2
2
m 2m + 3m 2 = (9x + 10)(9x – 10) = 4[(2z) – 5 ]
2
2
2 + 3m
= 4(2z + 5)(2z – 5)
FSTB/ HCF = 7m
Maka,
Thus,
7m(2 + 3m)
(b) 16y – 64y (e) 27m – 75 (h) 25a – 36
2
2
2
16 16y – 64y = 3(9m – 25) = (5a) – 6 2
2
2
2
y y – 4y = 3[(3m) – 5 ] = (5a + 6)(5a – 6)
2
2
2
y – 4
= 3(3m – 5)(3m + 5)
FSTB/ HCF = 16y
Maka,
Thus,
16y(y – 4)
(c) 15p q – 21pq 2 (f) 169u – 225 (i) 243g – 48g
3
2
2
3 15p q – 21pq 2 = (13u) – 15 2 = 3g(81g – 16)
2
2
2
p 5p q – 7pq 2 = (13u + 15)(13u – 15) = 3g[(9g) – 4 ]
2
2
2
q 5pq – 7q 2 = 3g(9g – 4)(9g + 4)
5p – 7q
FSTB/ HCF = 3pq
Maka,
Thus,
3pq(5p – 7q)

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