x^3-3x^2+3
Im not sure how to factor this
x^3-3x^2+3 = 0
let x = 1/y,
(1/y)^3 - 3(1/y)^2 + 3 = 0, MBS by y^3,
1 - 3y + 3y^3 = 0,
y^3 - y + 1/3 = 0. Very hard to factor by grouping.
y^3 = y - 1/3 = Ay + B, where A = 1 and B = -1/3
Let p = A/3 = 1/3
q = B/2 = (-1/3)/2 = -1/6
q^2 - p^3 = (-1/6)^2 - (1/3)^3 = 1/36 - 1/27 = -1/108(negative) < 0.
CASE III. q^2 - p^3 < 0, 3 roots are real and distinct. Determine an angle u between 0 and 180 degrees, such that cos u = q/(p^3/2) = (-1/6)/(1/3)^3/2 = -(1/2)(3)^(1/2)
u = 150 degrees or 210 will do but we choose 150 because of the restriction.
y1 = (2p^1/2) cos (u/3) = 2(1/3)^(1/2) cos (150/3) = 0.74228
y2 = (2p^1/2) cos (u/3 + 120) = 2(1/3)^(1/2) cos (150/3 + 120) = 2(1/3)^(1/2) cos (170) = 1.08310
y3 = (2p^1/2) cos (u/3 + 240) = 2(1/3)^(1/2) cos (150/3 + 240) = 0.39493
x1 = 1/y1 = 1/0.74228 = 1.3472
x2 = 1/y2 = 1/1.08310 = -0.8794
x3 = 1/y3 = 1/0.39493 = 2.532
See graph below.
By the "rational roots theorem", the only possible rational roots will be 1, -1, 3, or -3. (1)^2- 2(1)^2+ 3= 1- 2+ 3= 2, (-1)^3-2(-1)+ 3= -1- 2+ 3= 0. (3^3- 2(3^2)+ 3= 27- 18+ 3= 12, and (-3)^3- 3(-3)^2+ 3= -27- 3+ 3= 27.
So x= -1 is the only rational root. You can divide x^3- 3x^2+ 3 by x+ 1 to see that x^2 - 3x+ 3 is the other factor. Since its discriminant is negative, there is no further factorization (with integer coefficients).
x^3- 3x^2+ 3= (x+ 1)(x^2- 3x+ 3)