Factor: x^2+x+1 lol >_>;?

Question

Woot, forgot my basic math :(

The original was: (x^3-1) :O

Answer 1

Depending what you are trying to do, you could try:

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

If you are working in complex numbers, then you could also try:
x^2 +x +1 = {x + [1/2] + [(√-3)/2]}{x + [1/2] - [(√-3)/2]}


If you want to complete the square (and if you can get rid of the extra -x somewhere else) you can try:

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


However, in real numbers, x^2 +x +1 is prime (does not factor to lower degree).

---

Since it is a second degree, it can only factor to the form:

(x+a)(x+b) = x^2 + (a+b)x + ab

And there are, in real numbers, no combinations of a and b such that
ab=1 AND (a+b)=1

ab=1 implies that neither a nor b can be 0 (otherwise ab=0).

Can either one be +1?
No, because the other must then be zero (from a+b=1).

Can they both be negative?
No because then a+b would be negative (it must be +1).

Can they both be positive?
No because ab=1 implies that one of them must be greater than 1 (b = 1/a) and then a+b would be greater than 1.

We are left with the possibility that one is positive and one is negative?
No because then ab would be negative.

Therefore, the system
ab=1 and
a+b=1
has no real solution.

Answer 2

This polynomial cannot be factored. If you were to graph it, you would find that it only has positive values, and it never crosses or touches the x-axis. When a quadratic polynomial has no zeroes, that means it cannot be factored with real coefficients.

You are correct, however, that this is a factor of x^3 - 1.

Answer 3

x² + x + 1

Quadratic Formula:
[-b ± √(b² - 4ac]/ 2a
[-(1) ± √((1)² - 4(1)(1)]/ 2(1)
[-1 ± √(1 - 4]/ 2
[-1 ± √(- 3)]/ 2
[-1 ± √3(ì)]/ 2
[-1 ± 1∙732 050 808...ì]/ 2
x = -2∙732 050 808...ì/2 & x = 0∙732 050 808...ì/2
x = -1∙366 025 404...ì & x = 0∙36 025 403...ì

Answer 4

That cannot be factored in terms of real values.

Answer 5

It's prime.

Unless you've been taught the quadratic equation you should answer "prime."