Divide $2x^2 + 3x + 1$ by $x + 2$.

(N/A) To divide $2x^2 + 3x + 1$ by $x + 2$, we perform polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$ to get $2x$. This is the first term of the quotient.
$2$. Multiply the divisor $(x + 2)$ by $2x$ to get $2x^2 + 4x$. Subtract this from the dividend: $(2x^2 + 3x + 1) - (2x^2 + 4x) = -x + 1$.
$3$. Divide the first term of the new expression $(-x)$ by the first term of the divisor $(x)$ to get $-1$. This is the second term of the quotient.
$4$. Multiply the divisor $(x + 2)$ by $-1$ to get $-x - 2$. Subtract this from the current expression: $(-x + 1) - (-x - 2) = 3$.
Since the degree of the remainder ($3$, which is $0$) is less than the degree of the divisor ($x + 2$, which is $1$), we stop.
Thus, the quotient is $2x - 1$ and the remainder is $3$.
Verification:
$(2x - 1)(x + 2) + 3 = (2x^2 + 4x - x - 2) + 3 = 2x^2 + 3x + 1$.
Therefore, $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$.