How do you factor $8 x^{2} - 18$ $8x^2-18$ ?

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Answer 1 · 2018-04-04T19:00:57

$2 (2 x + 3) (2 x - 3)$ $2(2x+3)(2x-3)$

First, factor out the GCF (2)

$2 (4 x^{2} - 9) \to 4 x^{2} - 9$ $2(4x^2-9) rarr 4x^2-9$ can be factored out using difference of squares:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

$a$ $a$ in this case is $2 x$ $2x$ , $b$ $b$ is $3$ $3$

$(2 x - 3) (2 x + 3)$ $(2x-3)(2x+3)$

$2 (2 x - 3) (2 x + 3)$ $2(2x-3)(2x+3)$

How do you factor 8x^2-188x2−18 8x^2-18?