Question

Divide 4y^3-7y^2-3y-y by (2y+1) using remainder method

Answer 1

$\huge\mathfrak\red{Answer}$

$\frac{4y^{3} - 7y^{2} - 3y - y }{2y + 1} \\ = \frac{4y^{3} - 7y^{2} - 2y }{2y + 1} \\ = \frac{2y(2y^{2} - 7y - 1)}{2y + 1} \\ = \frac{2y^{2} - 7y - 1 }{1} \\ = 2y^{2} - 7y - 1$