Question

please provide the solution for the following
[y (cube) - 6y (square) - 40y ]divided by [ y(square) - 10y]

Answer 1

Answer:

y+4

Step-by-step explanation:

$\frac{y^3 - 6y^2 - 40y}{y^2 - 10y}\\= \frac{y(y^2 - 6y - 40)}{y(y - 10)}\\= \frac{y^2 - 6y - 40}{y - 10}\\= \frac{y^2 - 10y +4y - 40}{y - 10}\\= \frac{y(y - 10) +4(y - 10)}{y - 10}\\= \frac{(y - 10)(y + 4)}{y - 10}\\= y +4$