How do you factor #x^2+1/2x+1/4#?

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Answer 1 · 2018-03-02T03:42:25

The quadratic has complex factors: #(x-1/4(-1+sqrt(3)i)) and (x-1/4(-1-sqrt(3)i))#

#f(x) = x^2 + 1/2x +1/4#

Now, consider #f(x) =0#

Then if #a and b# are roots of #f(x) -> f(x) =(x-a)(x-b)#

#f(x)# is a quadratic with a discriminant of #1/4-4xx1xx1/4 =-3/4#

Since the discriminant #<0 -> a and b# will be complex.

#x= (-1/2 +-sqrt(-3/4))/2#

#= -1/4 +-1/4sqrt(3)i#

#= 1/4(-1+-sqrt(3)i)#

#:.# with #a and b# defined above:

#a = (x-1/4(-1+sqrt(3)i))#
and
#b= (x-1/4(-1-sqrt(3)i))#

Thus, #f(x)= (x-1/4(-1+sqrt(3)i))(x-1/4(-1-sqrt(3)i))#