Normally when you factor a polynomial, it is to make a problem easier. This is a polynomial of the form ax2 + bx - c, where a=b=c=1. Since a=1, to factor this polynomial nicely we would have to find two factors of c=1 that add up to b=1. Since the factors of 1 are -1 and 1, the possible numbers to add to are (-1) + (-1) = -2; 1 + (-1) = 0; and 1 + 1 = 2. None of these can add up to 1 (our b), so a nice answer is impossible.
However, when you factor, you are actually just finding solutions. For example, x2-3x+2= (x-1)(x-2). If x2-3x+2=0, then either x-1=0 or x-2=0, so x=1 or x=2.
Anything to the half power, is the square-root, so we write, using the quadratic formulas (b +/- (b2 - 4ac)1/2)/(2a), the solutions of x2 + x + 1 are (1 +/- (-3)1/2)/2. Since irepresents (-1)1/2, we can factor x2+x+1 = [x - 1/2 +i(31/2)/2] [x - 1/2 -i(31/2)/2].
(2x - 1)(5x + 8)
(x-1)(x-4)
(x + 2)(3x - 1)(3x + 1)
x^2-x+1 cannot be factorised
(x-3)(3x+1)
(7a - 1)(6a + 1)
There is no rational factorisation.
(5w + 1)(w + 1)
(6x - 1)(6x - 1)
(2x - 1)(5x + 8)
(x-1)(x-4)
(x-1)(x+2)
(x - 1)(5x + 7)
(x - 1)(2x - 9)
(b + 8)(b - 1)
(x-6)(x-1)
(2x-3)(x-1)