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

[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

 

[arildno]

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

 

[teng125]

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

 

[teng125]

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

 

[matt grime]

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

 

[HallsofIvy]

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.

 

[arildno]

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

 

[teng125]

oooo...okok thanks very much