Find the quotient and the remainder when $2x^2 - 7x - 15$ is divided by $2x + 3$.

(N/A) To divide $2x^2 - 7x - 15$ by $2x + 3$,we use polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.
$2$. Multiply the divisor $(2x + 3)$ by $x$: $x(2x + 3) = 2x^2 + 3x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + 3x) = -10x - 15$.
$4$. Divide the first term of the new expression $(-10x)$ by the first term of the divisor $(2x)$: $-10x / 2x = -5$. This is the second term of the quotient.
$5$. Multiply the divisor $(2x + 3)$ by $-5$: $-5(2x + 3) = -10x - 15$.
$6$. Subtract this from the current expression: $(-10x - 15) - (-10x - 15) = 0$.
Therefore,the quotient is $x - 5$ and the remainder is $0$.