Factorization for x^3 - 4x^2 -x = 0

1. Jul 12, 2006

teng125

for x^3 - 4x^2 -x = 0 , i have found one of the root which is 1 by dividing this equation by (x-1).
from there onwards i can't do already to find the other two roots.somebody pls help

thanx

2. Jul 12, 2006

arildno

1 is not a root, 0 is. Unless you have posted the wrong equation.

3. Jul 12, 2006

teng125

sorry.the matrix is [1 2 1;2 1 1;1 1 2].so i want to find the eigenvalues

4. Jul 12, 2006

teng125

therefore i got the eqn x^3 - 4x^2 -x = 0

5. Jul 12, 2006

matt grime

fine, but 1 is still not a root (1-4-1 is not zero)

6. Jul 13, 2006

HallsofIvy

Staff Emeritus
The characteristic equation for the matrix you give is x^3- 4x^2- x+ 4=0
not what you give. That equation does have 1 as a root so apparently you just wrote the equation wrong (twice!).
You say you found that 1 was a root "by dividing this equation by (x-1)" (Which makes me wonder why you chose x-1. It's simpler just to set x= 1 in the equation!). When you did that surely you found that
x^3- 4x^2- x+ 4= (x-1)(x^2- 3x- 4). Solve x^2- 3x- 4= 0. That factors easily but even it it didn't you could use the quadratic formula.

7. Jul 13, 2006

arildno

If you find the quadratic formula eerie, recognize that (1,1,1) is an eigenvector of your matrix.

8. Jul 13, 2006

teng125

oooo.....okok thanx very much