\(x=-2+\sqrt{5}>0\Rightarrow x+2=\sqrt{5}\)
\(\Rightarrow\left(x+2\right)^2=5\Rightarrow x^2+4x=1\)
Ta có:
\(3x^5+12x^4-8x^3-23x^2-7x+1\)
\(=3x^3\left(x^2+4x\right)-8x^3-23x^2-7x+1\)
\(=-5x^3-23x^2-7x+1=-5x\left(x^2+4x\right)-3x^2-7x+1\)
\(=-3x^2-12x+1=-3\left(x^2+4x\right)+1=-3+1=-2\)