Question

factorise 4(p+q)(3a-b)+6(p+q)(2b-3a)

Answer 1

4(p+q)(3a-b)+6(p+q)(2b-3a)=(p+q)[4(3a-b)+6(2b-3a)]
=(p+q)[12a-4b+12b-18a]=(p+q)(-6a+8b)
=-2(p+q)(3a-4b)

Answer 2

The factorisation of $4(p+q)(3a-b)+6(p+q)(2b-3a)$ is $2(p+q)(4b-3a).$

Step-by-step explanation:

We have,

$4(p+q)(3a-b)+6(p+q)(2b-3a)$

To find, the factorisation of $4(p+q)(3a-b)+6(p+q)(2b-3a)=?$

∴ $4(p+q)(3a-b)+6(p+q)(2b-3a)$

Taking common as 2(p + q), we get

$2(p+q)[2(3a-b)+3(2b-3a)]$

$=2(p+q)(6a-2b+6b-9a)$

$=2(p+q)(6a-9a-2b+6b)$

$=2(p+q)(-3a+4b)$

$=2(p+q)(4b-3a)$

Hence, the factorisation of $4(p+q)(3a-b)+6(p+q)(2b-3a)$ is $2(p+q)(4b-3a).$