b.
\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12x=0\)
\(\Leftrightarrow\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2+x+6=0\left(loại\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy.....
a)\(x^3+4x-5=x^3-x^2+x^2-x+5x-5=\left(x^2+x+5\right)\left(x-1\right)\)
b) \(x^4+2x^3+5x^2+4x-12=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=\left(x-1\right)\left(x^2+x+6\right)\left(x+2\right)\)
