\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2+4x-3x-12\right)\)
\(=\left(x-2\right)\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
Ta có:\(x^3-x^2-14x+24=\left(x^3-2x^2\right)+\left(x^2-2x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2-3x+4x-12\right)\)
\(=\left(x-2\right)\left[x\left(x-3\right)+4\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
Ta có:
\(x^3-x^2-14x+24\) \(=x^3+4x^2-5x^2-20x+6x+24\)
\(=x^2\left(x+4\right)-5x\left(x+4\right)+6\left(x+4\right)\)
\(=\left(x+4\right)\left(x^2-5x+6\right)\)
\(=\left(x+4\right)\left(x^2-3x-2x+6\right)\)
\(=\left(x+4\right)[x\left(x-3\right)-2\left(x-3\right)]\)
\(=\left(x+4\right)\left(x-2\right)\left(x-3\right).\)
Vậy \(x^3-x^2-14x+24=\left(x+4\right)\left(x-2\right)\left(x-3\right).\)
