If $f(x)=x^2+ax+b$ divides $f(x+2)f(x-2)$, find the minimum value of $f$. I am stuck with this Precalculus problem about polynomial functions. The problem:

Consider $f(x)=x^2+ax+b$ with $a^2-4b>0$. Let $\alpha$ and $\beta$ be the roots of $f$. Assume that $f(x)$ divides $f(x+2)f(x-2)$. Then

*

*Show that $\alpha=\beta+2$ or $\alpha=\beta-2$


*Using $1.$, find the minimum value of $f(x)$.

Part $1$ is easy: write $f(x+2)f(x-2)=f(x)g(x)$ and substitute $x=\alpha$ to obtain $f(\alpha+2)f(\alpha-2)=0$, so $\alpha+2$ or $\alpha-2$ is a root of $f$. This root cannot be $\alpha$, so $\alpha+2=\beta$ or $\alpha-2=\beta$.
I am stuck with part $2$. Any idea?
The minimum is obtained at $-\frac{a}{2}=\frac{\alpha+\beta}{2}=\beta+1$ or $\beta-1$ but I don't know how to continue.

$ f(x) = ( x - \alpha) ( x - \beta) = (x - c + 1)( x - c - 1) = (x-c)^2 - 1$
Hence, the minimum value is -1. 
