Question

4a2-9b2-16c2+24bc: factorise

Answer 1

The factor of $4a^2-9b^2-16c^2+24bc$ is $(2a+3b-4c)(2a-3b+4c)$

Step-by-step explanation:

Given: $4a^2-9b^2-16c^2+24bc$

Factor the polynomial

$\Rightarrow 4a^2-(9b^2+16c^2-24bc)$

$\Rightarrow 4a^2-[(3b)^2+(4c)^2-2\cdot 3b\cdot 4c]$

$\because 9a^2=(3a)^2, 16b^2=(4b)^2$

$\Rightarrow (2a)^2-(3b-4c)^2$

$\because x^2+y^2-2xy=(x-y)^2$

$[tex]\Rightarrow (2a+3b-4c)(2a-3b+4c)$

$\because x^2-y^2=(x+y)(x-y)$

The factor of $4a^2-9b^2-16c^2+24bc$ is $(2a+3b-4c)(2a-3b+4c)$

#Learn more:

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