Find the factors of the expression:$12{(2\mathrm{x} - 3\mathrm{y})}^2 - 16(3\mathrm{y} - 2\mathrm{x})$ 

The given expression: $12{(2\mathrm{x} - 3\mathrm{y})}^2 - 16(3\mathrm{y} - 2\mathrm{x})$ . 

First, we write the prime factors of each term, and we get:

$\begin{array}{l}\Rightarrow 12{\left(2\mathrm{x} - 3\mathrm{y}\right)}^2 = 2 \times 2 \times 3 \times \left(2\mathrm{x} - 3\mathrm{y}\right) \times \left(2\mathrm{x} - 3\mathrm{y}\right)\\ \\ \Rightarrow 16\left(3\mathrm{y} - 2\mathrm{x}\right) =-1 \times 2 \times 2 \times 2 \times 2 \times \left(2\mathrm{x} - 3\mathrm{y}\right)\end{array}$

Now, we will find the HCF as, 

$\begin{array}{l}\Rightarrow \mathrm{HCF}\left(12{\left(2\mathrm{x}-3\mathrm{y}\right)}^2,16\left(3\mathrm{y}-2\mathrm{x}\right)\right)=2\times 2\times \left(2\mathrm{x}-3\mathrm{y}\right)\\ \\ = 4\left(2\mathrm{x}-3\mathrm{y}\right)\end{array}$ 

Now, taking HCF common, from the above expression, we get:

$\begin{array}{l}= 12{\left(2\mathrm{x}-3\mathrm{y}\right)}^2-16\left(3\mathrm{y}-2\mathrm{x}\right)\\ \\ =4\left(2\mathrm{x}-3\mathrm{y}\right)\left(3\left(2\mathrm{x}-3\mathrm{y}\right)-\left(-4\right)\right)\\ \\ = \mathbf{4}\left(2x-3y\right)\left(6x-9y+4\right)\end{array}$

 
Hence, factors of $12{\left(2\mathrm{x}-3\mathrm{y}\right)}^2-16\left(3\mathrm{y}-2\mathrm{x}\right)=\mathbf{4}\left(2x-3y\right)\left(6x-9y+4\right)$ .