How can you factor x^2+x+1 completely ?
2 Answers
Given,
[ applying formula
Explanation:
Note that:
x^2+x+1
is in the standard form:
ax^2+bx+c
with
This has discriminant
Delta = b^2-4ac = color(blue)(1)-4(color(blue)(1))(color(blue)(1)) = -3
Since
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
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
Bonus
Note that:
(x-1)(x^2+x+1) = x^3-1
So the zeros
That is, they are cube roots of
They are often denoted:
omega = -1/2+sqrt(3)/2i" " and" "omega^2 = bar(omega) =-1/2-sqrt(3)/2i