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

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Answer 1 · 2016-07-06T16:26:47

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

To factor $2 x^{2} - 18$ $2x^2-18$

Begin by factoring out the 2 from each term

$2 (x^{2} - 9)$ $2(x^2-9)$

Now we recognize that $x^{2} - 9$ $x^2-9$ is the difference of two squares
$x \cdot x$ $x*x$ and $3 \cdot 3$ $3*3$

This factors to

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

How do you factor completely 2x2−182x^2-18?