# Thread: Solve for X values

1. ## Solve for X values

x^3-3x^2+3
Im not sure how to factor this

2. ## 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)

3. so factor for this eqn will be

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

4. 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.
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This cubic equation is hard to factor by grouping . . . .

5. 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!

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

Obviously, $x^2(x-3) + 3 \neq x^2(x-3) + 3(x-3)$ !!!

6. 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).

7. 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.

8. 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)