Math, asked by aaarohi, 11 months ago

Factorise : 9 (x+y)²- 16 (x - 3y)²​

Answers

Answered by DevyaniKhushi
1

{\text{Let\:(x + y) be a \: \& \:(x - 3y)\:be\:b }}

Now,

 {9a}^{2}  -  {16b}^{2}  \\  {(3a)}^{2}  -  {(4b)}^{2}  \\ (3a - 4b)(3a + 4b)

Putting the value of a & b in the term obtained above, we get,

{ \large{[3(x  + y) - 4(x - 3y)][3(x + y) + 4(x - 3y)]}} \\ \\  { \large{[(3x + 3y - 4x + 12y)][(3x + 3y + 4x - 12y)]}} \\  \\ { \large{[ - x + 15y][7x - 9y]}} \\  \\ { \large{ { - 7x}^{2} + 9xy + 105xy -  {135y}^{2}  }} \\  \\ { \red{ \large{ \boxed{{ - 7x}^{2}  + 114xy -  {135y}^{2} }}}}

Answered by Souviksvk10
6

Step-by-step explanation:

9 {(x + y)}^{2}  - 16 {(x - 3y)}^{2}  \\  =  {3}^{2}  {(x + y)}^{2}  -  {4}^{2}  {(x - 3y)}^{2}  \\  =  {(3x + 3y)}^{2} -  {(4x - 12y)}^{2}   \\  = (3x + 3y + 4x - 12y)(3x + 3y - 4x + 12y) \\  = (7x - 9y)(15y - x)

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