How do you factor the trinomial $x^2 + 4x+12$ ?

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Answer 1 · 2016-01-02T21:39:08

$x^2+4x+12 = (x+2-2sqrt(2)i)(x+2+2sqrt(2)i)$

$x^2+4x+12$ is of the form $ax^2+bx+c$ with $a=1$ , $b=4$ and $c=12$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 4^2-(4xx1xx12) = 16-48 = -32$

Since this is negative, $x^2+4x+12$ has no Real zeros.

It can still be factored using Complex coefficients.

Use the quadratic formula to find zeros:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)$

$=(-4+-sqrt(-32))/2 = (-4+-4sqrt(2)i)/2 = -2+-2sqrt(2)i$

Hence:

$x^2+4x+12 = (x+2-2sqrt(2)i)(x+2+2sqrt(2)i)$

How do you factor the trinomial x^2 + 4x+12 x^2 + 4x+12?