Question

(2x³+4x²-x+8) / (x² + 3x + 2)

Answer 1

$\dfrac{2x^3+4x^2 -x+8}{x^2 +3x+2}\\\\\\=\dfrac{2x^3+6x^2+4x-2x^2-6x-4+x+12}{}\\\\=\dfrac{2x(x^2 +3x+2)-2(x^2 +3x+2)+x+12}{x^2 +3x+2}\\\\\\=\dfrac{2x(x^2+3x+2)}{x^2 +3x +2 }- \dfrac{2(x^2 +3x +2)}{x^2 +3x +2 } + \dfrac{x+12}{x^2 +3x+2}\\\\\\=2x-2 + \dfrac{x+2}{x^2 +3x +2}\\\\\\=2x-2 + \dfrac{x+12}{x^2 +2x +x +2}\\\\\\=2x-2 + \dfrac{x+12}{x(x+2) + x+2}\\\\\\=2x-2 + \dfrac{x+12}{(x+1)(x+2)}$