# Solve for X values

• Oct 19th 2009, 03:51 PM
Nightasylum
Solve for X values
x^3-3x^2+3
Im not sure how to factor this
• Oct 19th 2009, 03:53 PM
Mycc82
Okay
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, 05:20 AM
jeremyparker10
so factor for this eqn will be

x= 3, and 2 imaginary fators will be there with "i"(iota)
• Oct 20th 2009, 08:02 AM
pacman
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, 08:47 AM
Defunkt
Quote:

Originally Posted by Mycc82
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)

This is wrong!

\$\displaystyle (x^2+3)(x-3) = x^2(x-3) + 3(x-3)\$ By the law of distributivity

Obviously, \$\displaystyle x^2(x-3) + 3 \neq x^2(x-3) + 3(x-3)\$ !!!
• Oct 20th 2009, 06:25 PM
mr fantastic
Quote:

Originally Posted by jeremyparker10
so factor for this eqn will be

x= 3, and 2 imaginary fators will be there with "i"(iota)

This is wrong. Please check the original question before making bold statements that rely on someone elses work (work which, as has been pointed out, is wrong).
• Oct 21st 2009, 01:03 AM
pacman
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, 02:27 AM
HallsofIvy
Quote:

Originally Posted by Nightasylum
x^3-3x^2+3
Im not sure how to factor this

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)