Page 8 - Praktis Strategi 2020 Matematik - Tingkatan 2

Matematik Tingkatan 2 Bab 2 Pemfaktoran dan Pecahan Algebra

7. Selesaikan. TP 2 (Buku Teks: m.s 27 – 28)
Solve.
Contoh

2
2
Diberi (3m + 2)(m – 1) = 3m – m – 2, senaraikan faktor-faktor bagi 3m – m – 2.
2
Given (3m + 2)(m – 1) = 3m – m – 2, list the factors of 3m – m – 2.
2
3m – m – 2 = (3m + 2)(m – 1)
2
= 1 × (3m + 2)(m – 1)
2
= (3m + 2) × (m – 1) (3m + 2)(m – 1) ialah hasil pemfaktoran 3m – m – 2.
(3m + 2)(m – 1) is the product of factors of 3m – m – 2.
2
2
Maka, faktor bagi 3m – m – 2 ialah 1, (3m + 2), (m – 1) dan (3m + 2)(m – 1).
Thus, the factors of 3m – m – 2 are 1, (3m + 2), (m – 1) and (3m + 2)(m – 1).
2
2
2
(a) Diberi (x + 5)(x – 3) = x + 2x – 15, senaraikan (b) Diberi (5 – 4q)(p + q) = 5p + 5q – 4pq – 4q ,
faktor-faktor bagi x + 2x – 15. senaraikan faktor-faktor bagi 5p + 5q – 4pq – 4q .
2
2
2
2
Given (x + 5)(x – 3) = x + 2x – 15, list the factors of Given (5 – 4q)(p + q) = 5p + 5q – 4pq – 4q , list the
x + 2x – 15. factors of 5p + 5q – 4pq – 4q .
2
2
2
2
x + 2x – 15 = 1 × (x + 5)(x – 3) 5p + 5q – 4pq – 4q = 1 × (5 – 4q)(p + q)
= (x + 5) × (x – 3) = (5 – 4q) × (p + q)
2
2
Maka, faktor bagi x + 2x – 15 ialah 1, (x + 5), Maka, faktor bagi 5p + 5q – 4pq – 4q ialah 1,
(x – 3) dan (x + 5)(x – 3). (5 – 4q), (p + q) dan (5 – 4q)(p + q).

8. Isi petak kosong di bawah. TP 1 (Buku Teks: m.s 27 – 28)
Fill in the blanks below.
Contoh
(a) 2 (4p + 3) = 8p + 6
6 (10y − 1 ) = 60y − 6
8p + 6 = 2(4p + 3)
Faktorkan 60y – 6.
60y – 6 = 6(10y – 1)
Factorise 60y – 6.

(b) 7 (8m + 3) = 56m + 21 (c) −5( 5 y – 4) = −25y + 20

56m + 21 = 7(8m + 3) −25y + 20 = −5(5y – 4)

TP 1 Mempamerkan pengetahuan asas tentang faktor.
TP 2 Mempamerkan kefahaman tentang konsep kembangan dan pemfaktoran.

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02 Bab 2.indd 16 11/12/19 12:28 PM