Question

12a²b- 9ab² + 6ab. factorise the following

Answer 1

Answer:

$\Large\boxed{3ab(4a-3b+2)}$

Step-by-step explanation:

Given:

$\Large\boxed{\textnormal{LESSON: FINDING THE FACTORISE!}}$

To find the factors of 12a²b-9ab²+6ab from left to right numbers. By using with distributive property.

Solutions:

First, thing you do is use exponent rule to expand the expression.

$\Large\boxed{\textnormal{EXPONENT RULES}}$

$\displaystyle A^B^+^C=A^B A^C$

$\displaystyle ABB=AB^2, AAB=A^2B$

Rewrite the whole problem down.

$\displaystyle 12AAB-9ABB+6AB$

Multiply the numbers from left to right. (Rewrite.)

$\displaystyle 3*2=6$

$\displaystyle 3*3=9$

$\displaystyle 3*4=12$

$\displaystyle 4*3AAB+3*3ABB+2*3AB$

Then, you have to solve with the factor of the common term of 3ab.

$\displaystyle 4*3AAB+3*3ABB+2*3AB=\boxed{3ab(4a-3b+2)}$

$\Large\boxed{3ab(4a-3b+2)}$

Hence, the correct answer is 3ab(4a-3b+2).

Answer 2

Answer:

$\mathrm{Factor}\:12a^2b-9ab^2+6ab:\quad 3ab\left(4a-3b+2\right)$

Step-by-step explanation:

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

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c}$

$\gray{ab^2=abb,\:a^2b=aab}$

$=12aab-9abb+6ab$

$\gray{\mathrm{Rewrite\:}6\mathrm{\:as\:}2\cdot \:3}$

$\gray{\mathrm{Rewrite\:}-9\mathrm{\:as\:}3\cdot \:3}$

$\gray{\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3}$

$=4\cdot \:3aab+3\cdot \:3abb+2\cdot \:3ab$

$\gray{\mathrm{Factor\:out\:common\:term\:}3ab}$

$=3ab\left(4a-3b+2\right)$