# How do you find the quotient of (14y^5+21y^4-6y^3-9y^2+32y+48)div(2y+3) using long division?

Feb 13, 2018

Quotient is $7 {y}^{4} - 3 y + 2 + 16$

#### Explanation:

$14 {y}^{5} + 21 {y}^{4} - 6 {y}^{3} - 9 {y}^{2} + 32 y + 48$

=$7 {y}^{4} \cdot \left(2 y + 3\right) - 3 y + 2 \cdot \left(2 y + 3\right) + 16 \cdot \left(2 y + 3\right)$

=$\left(2 y + 3\right) \cdot \left(7 {y}^{4} - 3 y + 2 + 16\right)$

Hence quotient is $7 {y}^{4} - 3 y + 2 + 16$ and remainder is $0$.