How do you factor the trinomial $x^{2} - 4 x - 12$ $x^2 - 4 x - 12$ ?

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Answer 1 · 2016-05-03T12:40:05

$x^{2} - 4 x - 12 = (x - 6) (x + 2)$ $x^2-4x-12 = (x-6)(x+2)$

Find a pair of factors of $12$ $12$ which differ by $4$ $4$ .

The pair $6, 2$ $6, 2$ works, in that $6 \cdot 2 = 12$ $6*2=12$ and $6 - 2 = 0$ $6-2=0$

Hence:

$x^{2} - 4 x - 12 = (x - 6) (x + 2)$ $x^2-4x-12 = (x-6)(x+2)$

How do you factor the trinomial x^2 - 4 x - 12x2−4x−12 x^2 - 4 x - 12?