Question

Factorise 4 a square minus b square minus 16 C square + 24 BC

Answer 1

The factorisation of $4a^{2}-9b^{2}-16c^{2} +24bc$ is $(2a+3b-4c)(2a-3b+4c)$.

Step-by-step explanation:

We have,

$4a^{2}-9b^{2}-16c^{2} +24bc$

To find, the factorisation of $4a^{2}-9b^{2}-16c^{2} +24bc=?$

∴ $4a^{2}-9b^{2}-16c^{2} +24bc$

$=4a^{2}-(9b^{2}+16c^{2}-24bc)$

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

$=4a^{2}-(3b-4c)^2$

Using algebraic identity,

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

$=(2a)^{2}-(3b-4c)^2$

$=(2a+3b-4c)(2a-3b+4c)$

Using algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

Hence, the factorisation of $4a^{2}-9b^{2}-16c^{2} +24bc$ is $(2a+3b-4c)(2a-3b+4c)$.