有一條因式分解我5識...可5可以教我,,,我明天要測驗..

(1)已知3pq-2q=-5,求12pq-8q的值.
12pq-8q
= 4(3pd-2q)
=4(-5)
=-20

(2a)因式分解2a^2b^2-4ab.
2a^2b^2-4ab
= 2ab(ab-2)

(2b)因式分解3ab-6
3ab-6
=3(ab-2)

(2c)運用(a)和(b)的結果,因式分解2a^2b^2-ab-6
2a^2b^2-ab-6
=2a^2b^2-4ab+4ab-ab-6 <--因想做到a,b的樣子, 所以+4ab,再-4ab
=(2a^2b^2-4ab)+(3ab-6) <--by the result of a & b
=2ab(ab-2)+3(ab-2)
=(2ab-3)(ab-2)