# How do find the quotient of (6x^3+5x^2+2x+1)/ (2x+3)?

$6 {x}^{3} + 5 {x}^{2} + 2 x + 1 = \left(2 x + 3\right) \left(3 {x}^{2} - 2 x + 4\right) - 11$
Suppose that $q \left(x\right)$ is the quocient and $n \left(x\right) = 6 {x}^{3} + 5 {x}^{2} + 2 x + 1 , d \left(x\right) = 2 x + 3 , r \left(x\right) = {c}_{0}$. Then $q \left(x\right) d \left(x\right) + r \left(x\right) = n \left(x\right)$ Putting $q \left(x\right) = a {x}^{2} + b x + c$ (Here $q \left(x\right)$ degree must be $3 - 1$ because $d \left(x\right)$ degree is $1$. By the same reason $r \left(x\right)$ degree is $0$).
After multiplication of $q \left(x\right) d \left(x\right) + r \left(x\right) - n \left(x\right) = 0$ we get:
$\left(2 a - 6\right) {x}^{3} + \left(3 a + 2 b - 5\right) {x}^{2} + \left(3 b + 2 c - 2\right) x + 3 c + {c}_{0} - 1 = 0$. This equation must be identically null for all $x$. Solving for $a , b , c , {c}_{0}$ we obtain:
$q \left(x\right) = 3 {x}^{2} - 2 x + 4 , r \left(x\right) = - 11$