Computing Minima for degree 8 polynomial

Question

Is there any better method than derivative or trial and error to calculate minima of $$f(x)=x^8-8x^6+19x^4-12x^3+14x^2-8x+9$$

The minima occurs at $x=2$ and $f(2)=1$

We were given option too $-1,9,6,1$ if that helps

Answer 1

By splitting conveniently the terms (by completing squares from left to right) we obtain $$\begin{align}f(x)&=x^8-8x^6+(16+3)x^4-12x^3+(12+2)x^2-8x+(8+1) \\&=(x^8-8x^6+16x^4)+(3x^4-12x^3+12x^2)+(2x^2-8x+8)+1\\ &=x^4(x^2-4)^2+3x^2(x-2)^2+2(x-2)^2+1\geq 1=f(2).\end{align}$$