Question

Factorize : 4a² − 9b² − 16c² + 24bc

Answer 1

Here is the identity you need, simple but important at the same time.

→→$a^2-b^2=(a+b)(a-b)$

Let's complete the square first.

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

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

Now the the identity can be applied!

∴$(2a+3b-4c)(2a-3b+4c)$ ←← ANSWER

Answer 2

Factors are:

$4a^2-9b^2-16c^2+24bc$ = $(2a-3b+4c)\ and\ (2a+3b-4c)$

Explanation:

In this question, it is required to factorize the given expression :

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

For this, firstly it is required to simplify the given expression as :

$(2a)^2-(3b)^2-(4c)^2+2(3)(4)bc\\\\=(2a)^2-[(3b)^2+(4c)^2-2(3)(4)bc]$...........(1)

We know that,

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

Equation (1) becomes :

$(2a)^2-[(3b)^2+(4c)^2-2(3)(4)bc]\\\\=(2a)^2-(3b-4c)^2$

Since, $a^2-b^2=(a-b)(a+b)$

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

So, the factors of $4a^2-9b^2-16c^2+24bc$ are $(2a-3b+4c)\ and\ (2a+3b-4c)$.