How do you factor $3 x^{2} - 12$ $3x^2-12$ ?

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Answer 1 · 2018-03-21T04:09:48

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

First, we find any like terms that we can factor out of the entire equation. In this case, we can factor out a three.

This gives us:
$3 (x^{2} - 4)$ $3(x^2-4)$

We can simplify this even further using The Difference of Squares

$x^{2} - 4 = 0$ $x^2-4 = 0$
$x^{2} = 4$ $x^2 = 4$
$\sqrt{x^{2}} = \sqrt{4}$ $sqrt(x^2) = sqrt(4)$
$x = 4, - 4$ $x = 4, -4$

Therefore the factors of $x^{2} - 4$ $x^2-4$ are $(x - 2)$ $(x-2)$ and $(x + 2)$ $(x+2)$

Putting this all together, we get:

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

How do you factor 3x2−123x^2-12?