Find the quotient and the remainder when $2x^2 - 7x - 15$ is divided by $x + 1$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $2x^2 - 7x - 15$ by $x + 1$.
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$: $2x^2 / x = 2x$. This is the first term of the quotient.
$2$. Multiply the divisor $(x + 1)$ by $2x$: $2x(x + 1) = 2x^2 + 2x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + 2x) = -9x - 15$.
$4$. Divide the first term of the new expression $(-9x)$ by the first term of the divisor $(x)$: $-9x / x = -9$. This is the second term of the quotient.
$5$. Multiply the divisor $(x + 1)$ by $-9$: $-9(x + 1) = -9x - 9$.
$6$. Subtract this from the current expression: $(-9x - 15) - (-9x - 9) = -9x - 15 + 9x + 9 = -6$.
Thus,the quotient is $2x - 9$ and the remainder is $-6$.