Question

6\(x^{3}\) - 17\(x^{2}\) - 4x + 3=0

Answer 1

\(\Leftrightarrow6x^3-18x^2+x^2-3x-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^2+x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)\left(3x-1\right)=0\)

hay \(x\in\left\{3;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

Answer 2

\(\Leftrightarrow6x^3-18x^2+x^2-3x-x+3=0\)

\(\Leftrightarrow6x^2\left(x-3\right)+x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)