The factorization of the expression [tex]9 x^{2}-4 y^{2}[/tex] is (3x+2y) (3x-2y).
Solution:
In the expression [tex]9 x^{2}-4 y^{2}, 9 x^{2}[/tex] can be written as [tex](3 x)^{2}[/tex] . Similarly [tex]4 y^{2}[/tex] can be written as [tex](2 y)^{2}[/tex]
[tex]9 x^{2}-4 y^{2}=(3 x)^{2}-(2 y)^{2}[/tex]
Since both terms [tex](3 x)^{2}[/tex] and [tex](2 y)^{2}[/tex] are perfect squares, using the difference of squares formula,
[tex]a^{2}-b^{2}=(a+b)(a-b)[/tex]
Here a = (3x) and b = (2y)
[tex](3 x)^{2}-(2 y)^{2}=(3 x+2 y)(3 x-2 y)[/tex]
(3x+2y) and (3x-2y) are the factors of [tex]9 x^{2}-4 y^{2}[/tex]