Question #462d9

Question

2113 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-25T16:06:57

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

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

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

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.

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

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

Replace p and q by their real values to get:

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