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)#