Question

(40x² – 20x³ + 10xy) : (–10x)

Answer 1

    $\frac{40x^2-20x^3+10xy}{-10x}=$

$=\frac{-20x^3+40x^2+10xy}{-10x}=$

$=\frac{-20x^2+40x+10y}{-10}=$

$=\frac{20x^2-40x-10y}{10}=$

$=2x^2-4x-y$