Use the quadratic formula as there are no integer roots.
x2 + 3x + 2 factors as (x + 1)(x + 2) and x2 - 3x + 2 factors as (x - 1)(x - 2)
(x + 2)(x - 1)
x2 + 5x + not really sure what that extra plus is for... this is the most you can factor it... x(x + 5)
x - 4x + 4 = -3x + 4 which cannot be factorised.
-x(x2 - 2)(x2 + 3)
Take out the common factor, 3: 3x + 6 = 3(x + 2).
It can be. x^2 + x + 1 is a factor of 2x^2 + 2x + 2
(x + 2)(x + 1)
2+2y+x+xy=(x+2)(y+1)
x(x^2 + 1)
x^3 + x^2 + x = x(x^2 + x + 1)
(x - 1)(x - 2)
(x + 2)(x - 1)
(x + 1)(x^2 - x + 1)
The factors of x^3 - x^2 + 2 are (x + 1) and (x^2 - 2x + 2)
(2x + 1)(x + 2)
-1
3(x + 1)(x + 2)