# Factoring a symbolic polynomial

The following code:

```
llambda, mu = var('λ, μ')
uVars = list(var(', '.join([f'u{n}' for n in range(1, 3 + 1)])))
aVars = list(var(', '.join([f'a{n}' for n in range(1, 3 + 1)])))
U = Matrix([[0, -uVars[2], uVars[1]],
[uVars[2], 0, -uVars[0]],
[-uVars[1], uVars[0], 0]])
a = Matrix([[aVars[0], 0, 0],
[0, aVars[1], 0],
[0, 0, aVars[2]]])
I = matrix.identity(3)
L = a*llambda + U
# Characteristic polynomial
charPoly = det((L - mu*I))
factor(charPoly)
```

computes a characteristic polynomial and yields:

```
a1*a2*a3*λ^3 - a1*a2*λ^2*μ - a1*a3*λ^2*μ - a2*a3*λ^2*μ
+ a1*u1^2*λ + a2*u2^2*λ + a3*u3^2*λ + a1*λ*μ^2
+ a2*λ*μ^2 + a3*λ*μ^2 - u1^2*μ - u2^2*μ - u3^2*μ - μ^3
```

However, this is not the simplification I desire. I want this: $$(a_1\lambda - \mu)(a_2\lambda - \mu)(a_3\lambda - \mu) - (u_1^2 + u_2^2 + u_3^2)\mu + (a_1 \mu_1^2 + a_2 u_2^2 + a_3 u_3^2)\lambda$$

Is there a way to obtain that sort of factorization?

I don't think there is an easy way to achieve this. Even Mathematica doesn't know what to do with this expression (If you have it installed, you can try with

`charPoly._mathematica_().Factor()._sage_()`

, but you need to use plain text variables). Sometimes it is a bit more powerful than Maxima, but not in this case.