1.
\(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+5x+x+5=0\)
\(\Leftrightarrow\left(2x^3+2x^2\right)+\left(x^2+x\right)+\left(5x+5\right)=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0\) hoặc \(2x^2+x+5=0\)
mà \(2x^2+x+5>0\)
Do đó x = -1