Question

factorise the following 12a^2b-9ab^2+6ab

Answer 1

Answer.


$12 {a}^{2} b - 9a {b}^{2} + 6ab$


$3(4 {a}^{2} b - 3a {b}^{2} + 2ab) = 0$



$4 {a}^{2}b - 3a {b}^{2} + 2ab = 0$


$ab(4a - 3b + 2) = 0$




so, factores are


$ab \: \: \: \: \: \: (4a - 3b + 2)$

Answer 2

Here your Answer✍️

Your Question
➡️12a²b-9ab²+6ab

We can Common =3ab
3ab×4a=12a²b
3ab×-3b=-9ab²
3ab×2=6ab

So,
3ab(4a-3b+2)

Factors=3ab(4a-3b+2)

hOpe you understand