x^3-3x^2+3

Im not sure how to factor this

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- Oct 19th 2009, 04:51 PMNightasylumSolve for X values
x^3-3x^2+3

Im not sure how to factor this - Oct 19th 2009, 04:53 PMMycc82Okay
So, first would probably be grouping.

You would take the x^3 and group it with the 3x^2.

Factor that into x^2 (x-3) + 3

Then it would just be (x^2+3) (x-3) - Oct 20th 2009, 06:20 AMjeremyparker10
so factor for this eqn will be

x= 3, and 2 imaginary fators will be there with "i"(iota) - Oct 20th 2009, 09:02 AMpacman
you have 3 real roots, see the graph. By Descartes law of signs, (+) 2 sign changes, (-) only 1. So we have two positive roots and one negative root.

hhttp://www2.wolframalpha.com/Calcula...image/gif&s=23

(Bow)

This cubic equation is hard to factor by grouping . . . . - Oct 20th 2009, 09:47 AMDefunkt
- Oct 20th 2009, 07:25 PMmr fantastic
- Oct 21st 2009, 02:03 AMpacman
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. (Bow) - Oct 21st 2009, 03:27 AMHallsofIvy
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)