How can you factor #x^2+x+1# completely ?

Given, #x^2+x+1#

#rArr x^2+2x+1-x#

#rArr (x+1)^2-(sqrt x)^2 #[ formula #a^2+2ab+b^2 = (a+b)^2]#

#rArr (x+1+sqrt x)(x+1- sqrt x)#
[ applying formula #a^2 -b^2 = (a+b)(a-b)#]

Note that:

#x^2+x+1#

is in the standard form:

#ax^2+bx+c#

with #a=1#, #b=1# and #c=1#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = color(blue)(1)-4(color(blue)(1))(color(blue)(1)) = -3#

Since #Delta < 0# this quadratic has no real zeros and no linear factors with real coefficients.

We can still factor it, but we need to use non-real Complex coefficients.

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

We can complete the square then use this with #A=(x+1/2)# and #B=sqrt(3)/2i# as follows:

#x^2+x+1 = x^2+x+1/4+3/4#

#color(white)(x^2+x+1) = (x+1/2)^2+(sqrt(3)/2)^2#

#color(white)(x^2+x+1) = (x+1/2)^2-(sqrt(3)/2i)^2#

#color(white)(x^2+x+1) = ((x+1/2)-sqrt(3)/2i)((x+1/2)+sqrt(3)/2i)#

#color(white)(x^2+x+1) = (x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)#

where #i# is the imaginary unit, satisfying #i^2=-1#

Bonus

Note that:

#(x-1)(x^2+x+1) = x^3-1#

So the zeros #-1/2+-sqrt(3)/2i# of #(x^2+x+1)# that we found above are also zeros of #(x^3-1)#.

That is, they are cube roots of #1#.

They are often denoted:

#omega = -1/2+sqrt(3)/2i" "# and #" "omega^2 = bar(omega) =-1/2-sqrt(3)/2i#

#omega# is called the primitive complex cube root of #1# and crops up a lot when solving cubic equations lacking simple roots.