# Question 462d9

Jun 25, 2016

$3 \left[\left(9 \left(3 x - 1\right) + 4 \left(2 y + 3\right)\right) \left(9 \left(3 x - 1\right) - 4 \left(2 y + 3\right)\right)\right]$

#### Explanation:

Always look for a common factor first.  is a common factor here.

$243 {\left(3 x - 1\right)}^{2} - 48 {\left(2 y + 3\right)}^{2}$
$= 3 \left[81 {\left(3 x - 1\right)}^{2} - 16 {\left(2 y + 3\right)}^{2}\right]$

Inside the brackets we have the difference of 2 squares.
This will be easier to recognise if we let (3x-1) = p
and (2y+3) = q.

$3 \left[81 {\left(3 x - 1\right)}^{2} - 16 {\left(2 y + 3\right)}^{2}\right]$ becomes
$3 \left[81 {p}^{2} - 16 {q}^{2}\right]$

=$3 \left[\left(9 p + 4 q\right) \left(9 p - 4 q\right)\right]$

Replace p and q by their real values to get:

$3 \left[\left(9 \left(3 x - 1\right) + 4 \left(2 y + 3\right)\right) \left(9 \left(3 x - 1\right) - 4 \left(2 y + 3\right)\right)\right]$