How to factor x^2+x+1?

Question

How to factor x^2+x+1?

Answer 1

I don't think it is possible to factor because the two numbers that multiply to 1 are 1 and -1.

1x1 = 1 but 1+1 = 2 not 1

-1x-1 = 1 but -1 + -1 = -2 not 1

Therefore

(x+1)(x+1) and (x-1)(x-1) don't work

Answer 2

x² + x + 1 =

since it has no real zeros, it can't be factored in the usual way;

nevertheless, you can rewrite it factored resorting to the following artifice:

rewrite x as 2x - x:

x² + 2x - x + 1 =

group the terms as:

(x² + 2x + 1) - x  =

(x + 1)² - x  =

now, try to view it as a difference between two squares, that is:

(x + 1)² - (√x)² =

thus factorable as:

[(x + 1) + (√x)] [(x + 1) - (√x)] =

(x + 1 + √x) (x + 1 - √x)

in conclusion:

x² + x + 1 = (x + √x + 1) (x - √x + 1)

I hope it helps...

Bye!

Answer 3

x^2 + x + 1 does not factor over the reals

because (x + 1/2)^2 = x^2 + x + 1/4

which means that for x^2 + x + C to have real zeros, C must be less than or equal to 1/4.

If you really need to factor it, however, you just need

(x - (first solution))(x - (second solution)).

Since someone else already gave you the complex solutions, I don't care to repeat them. (nor even to check them)

Answer 4

It doesn't factor over whole numbers.

Answer 5

factor x² + x + 1

You cannot factor this.

If it were set equal to zero, you can solve using the quadratic formula to solve for x which is:

[-b ± √(b² - 4ac)] / 2a where a = 1, b = 1, and c = 1

[-1 ± √(1² -4*1*1)] / 2*1

[-1 ± √(1-4)] / 2

[-1 ± √(-3)] / 2