Question

factorise 4a2-9b2-16c2+24bc​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

Step-by-step explanation:

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

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

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

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

By using identity:$(a-b)^2=a^2+b^2-2ab$

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

Using identity:$a^2-b^2=(a+b)(a-b)$

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

Hence, the factors of $4a^2-9b^2-16c^2+24bc$ are

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

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