How do you factor completely $4 x^{2} - 9 y^{2}$ $4x^2 - 9y^2$ ?

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Answer 1 · 2015-11-17T10:15:45

$(2 x + 3 y) (2 x - 3 y)$ $color(blue)((2x+3y)(2x-3y)$ is the factorised form of the expression

$4 x^{2} - 9 y^{2}$ $4x^2-9y^2$ can be also written as ${(2 x)}^{2} - {(3 y)}^{2}$ $(2x)^2-(3y)^2$

Now, as per identity:
$a^{2} - b^{2} = (a + b) (a - b)$ $color(blue)(a^2-b^2=(a+b)(a-b)$

So , in case of our expression
$a = 2 x, b = 3 y$ $color(blue)(a=2x, b=3y$

${(2 x)}^{2} - {(3 y)}^{2} = (2 x + 3 y) (2 x - 3 y)$ $(2x)^2-(3y)^2=color(blue)((2x+3y)(2x-3y)$

How do you factor completely 4x2−9y24x^2 - 9y^2?