How do you factor #x^2+x+3#?

Question

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Answer 1 · 2015-04-12T22:28:49

This expression has no real factors

But if you really want to factorize you can obtain complex factors

Method

Find the roots using the quadratic formula

#x = (-1 +- srqt(1 - 14))/2#

#=> x = -1/2 +- sqrt(-13)/2#

#=> x = -1/2 +- (13i)/2#

Therefore,

#x = -1/2 + (13i)/2 => x + 1/2 - (13i)/2 = 0#
and
#x = -1/2 - (13i)/2 => x + 1/2 + (13i)/2 = 0#

Hence,

#(x + 1/2 - (13i)/2)(x + 1/2 + (13i)/2) = 0#

Therefore Factorizing #x^2 + x + 3 = (x + 1/2 - (13i)/2)(x + 1/2 + (13i)/2)#