Factoring 3rd degree polynomial for eigenvalues

[hops1]

Homework Statement

Was given a matrix
To find the eigenvalues I set up the characteristic equation
[-1-x | 7 | -5 ]
[-4 | 11-x | -6 ]
[-4 | 8 | -3-x]

With some dirty work I got this bad boy out, which I'm having trouble factoring
-x³+7x²-15x+9

Homework Equations

Looking for method to factor it, without aid of calculator
Or if there's a better way to determine the equation which will give me the factors without setting up a polynomial

The Attempt at a Solution

I tried the grouping method but that doesn't work since all I end up with is
-x²(x-7)+3(5x-3)

 

[Ray Vickson]

Homework Statement

Was given a matrix
To find the eigenvalues I set up the characteristic equation
[-1-x | 7 | -5 ]
[-4 | 11-x | -6 ]
[-4 | 8 | -3-x]

With some dirty work I got this bad boy out, which I'm having trouble factoring
-x³+7x²-15x+9

Homework Equations

Looking for method to factor it, without aid of calculator
Or if there's a better way to determine the equation which will give me the factors without setting up a polynomial

The Attempt at a Solution

I tried the grouping method but that doesn't work since all I end up with is
-x²(x-7)+3(5x-3)

Use the rational root theorem on the polynomial $x^3 - 7x^2 + 15 x - 9$; that is, look for roots among the + or - integer factors of -9.

 

[hops1]

Use the rational root theorem on the polynomial $x^3 - 7x^2 + 15 x - 9$; that is, look for roots among the + or - integer factors of -9.

Thanks I forgot about that theorem.
But how do I rule out the negative or positive integers?

 

[Mark44]

You can rule out a potential root r by discovering that x - r is not a factor of your cubic polynomial. The rational root theorem says that the possible rational roots for your cubic are 1, -1, 3, -3, 9, and -9.

Use either long division or synthetic division to determine whether x - 1, x - (-1), x - 3, x - (-3), x - 9, or x - (-9) are factors. If one of these is a factor, the remainder will be zero. If it's not a factor, the remainder will be nonzero.

 

[hops1]

You can rule out a potential root r by discovering that x - r is not a factor of your cubic polynomial. The rational root theorem says that the possible rational roots for your cubic are 1, -1, 3, -3, 9, and -9.

Use either long division or synthetic division to determine whether x - 1, x - (-1), x - 3, x - (-3), x - 9, or x - (-9) are factors. If one of these is a factor, the remainder will be zero. If it's not a factor, the remainder will be nonzero.

Thanks

 

[Ray Vickson]

Thanks I forgot about that theorem.
But how do I rule out the negative or positive integers?

In this case you can put x = -y to find that the polynomial has the form p(x) = -q(y), where $q(y) = y^3 + 7 y^2 + 15 y +9$. Obviously, q(y) does not have any positive roots (all its coefficients are > 0), so p(x) does not have any negative roots. This type of trick does not always work, but it happens to be OK in this example. Even if it did not work, you could just try out all the factors of -9 to see if one of them sets p(x) = 0. As soon as you find one that works you can stop checking and start again with the resulting quadratic remaining factor.