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

2 Answers
Oct 2, 2017

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)]

Oct 2, 2017

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

Explanation:

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.