\(x^3+6x^2+12x-19=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\\ \Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\\ \Leftrightarrow x-1=0 \Leftrightarrow x=1\)
Vì: \(x^2+7x+19=\left(x^2+2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)+\dfrac{27}{4}=\left(x+\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)
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