Question

x^3+3x^2-4x-12=0

 

Answer 1

x = -3 hoặc x = 2 hoặc x = -2

Answer 2

\(\Leftrightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow x\)

Answer 3

\(x^3+3x^2-4x-12=0\)

\(\Leftrightarrow x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+5x+6\right)=0\)

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

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

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

Answer 4

\(x^3+3x^2-4x-12=0\)

\(x^2\left(x+3\right)-4\left(x+3\right)=0\)

\(\left(x^2-4\right)\left(x+3\right)=0\)

\(\left[{}\begin{matrix}x-2=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-3\end{matrix}\right.\)