Linear Algebra - minimal polynomial, polynomial

Question

the minimal polynomial of $A$ is $(x−1)(x+1)$. Let $f(x)=4x^{2008} − 8x^{597} + 10x + 6$

1. show $f(A) = \alpha I + \beta A$

2. $\alpha=?\ \beta=?$

So I worked on a bit, and I got this far

$A = x^2 - 1$

$A^2 = x^4 - 2x^2 + 1$

Any help would be appreciated.

Answer 1

Since $A^2-I=0$, you have $\,f(A)=(f\bmod x^2-1)(A)$.

Now, since $x^2\equiv 1\mod x^2-1$, we have: \begin{align*}f(x)\bmod x^2-1&=4(x^{2008}\bmod x^2-1)-8(x^{597}\bmod x^2-1)+10x+6\\ &=4-8x+10x+6=2x+10,\end{align*} whence $\,\alpha=10,\enspace \beta=2$.