Verified answer

x^3 – 6x^2+ 9x+6 =

x^3 – 6x^2+ 12x - 8 - 3x + 6 + 8 =

(x-2)^3 -3(x-2) + 8 =0

x-2 = t

t^3 - 3t +8 =0

t = u+v

(u+v)^3 - 3uv(u+v) -u^3 -v^3 =0 it's an identity

then

-3uv = -3 then u = 1/v

8 = -u^3-v^3

u^3+1/u^3+8 = 0

(u^3)^2 +8(u^3) +1 =0

u^3 = -4 ± √(16-1) = -4±√(15)

u = (-4+√(15))⅓

v = 1/u = (-4+√(15))(-⅓)

t = u+v = (-4+√(15))⅓ + (-4+√(15))(-⅓)

x = 2 +t

x = 2+ (-4+√(15))⅓ + (-4+√(15))(-⅓)

First factor

(x - 2 - (-4+√(15))⅓ - (-4+√(15))(-⅓))

The second factor it's a polinomial of two degree polynomial

which has complex root

From

u^3 = -4±√(15) obtain another two roots

u = [- 1/2 + i (√3)/2 ]*(-4+√(15))⅓

and

u = [- 1/2 - i (√3)/2 ]*(-4+√(15))⅓

find v = 1/ u , t = u+v, x = 2+t and you get factors over C, complex

The real root of x^3 – 6x^2+ 9x+6 = 0

it's x = - 0.492033301171817.....

exact value x = 2+ (-4+√(15))⅓ + (-4+√(15))(-⅓)

this is prime under rational numbers...

it has one real (irrational) root, and two complex conjugate roots...

edit:

f(x) = x^3 - 6x^2 + 9x + 6

f ' (x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3)

f ' (x) = 3(x - 3)(x - 1) ==> local extrema at x = 1 and x = 3

f " = 6x - 12

f " (1) = 6 - 12 = -6 ==> (1 , 11) is a local max

f " (3) = 6(3) - 12 = 6 ==> (3 , 6) is a local min

(f ( 3) = 27 - 6(9) + 27 + 6 = 6 )

f " = 0 at x = 2, so (2 , 8) is a point of inflection

x^3 – 6x^2 + 9x + 6

= (x + 4)(x^2 + 3x + 2)