What is the factors for $4 x^{2} - 9$ $4x^2 -9$ ?

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Answer 1 · 2015-05-06T04:46:47

$4 x^{2} - 9 = {(2 x)}^{2} - 3^{2}$ $4x^2 - 9 = (2x)^2 - 3^2$

We know that $a^{2} - b^{2} = (a + b) (a - b)$ $color(blue)(a^2 - b^2 = (a+b)(a-b)$

Hence ${(2 x)}^{2} - 3^{2} = (2 x + 3) (2 x - 3)$ $(2x)^2 - 3^2 = (2x + 3)(2x - 3)$

The factors of $4 x^{2} - 9$ $4x^2 - 9$ are $(2 x + 3)$ $color(green)((2x + 3)$ and $(2 x - 3)$ $color(green)((2x - 3)$

Answer 2 · 2015-05-06T04:47:17

The answer is $(4 x^{2} - 9) = (2 x - 3) (2 x + 3)$ $(4x^2-9)=(2x-3)(2x+3)$ .

This is an example of the difference of squares, where $(a^{2} - b^{2}) = (a - b) (a + b)$ $(a^2-b^2)=(a-b)(a+b)$ .

$(4 x^{2} - 9) = ({(2 x)}^{2} - 3^{2})$ $(4x^2-9)=((2x)^2-3^2)$

$a = 2 x$ $a=2x$
$b = 3$ $b=3$

$(4 x^{2} - 9) = (2 x - 3) (2 x + 3)$ $(4x^2-9)=(2x-3)(2x+3)$

What is the factors for 4x^2 -94x2−94x^2 -9?