1. **State the problem:** We are given that $x = -2$ is a root of the polynomial equation $$15x^3 + 26x^2 - 11x - 6 = 0.$$ We need to find values of $p$ and $q$ such that $$15x^3 + 26x^2 - 11x - 6$$ is a factor of $$15x^4 + px^3 - 37x^2 + qx + 6.$$\n\n2. **Understand the problem:** Since the cubic polynomial is a factor of the quartic polynomial, there exists a linear polynomial $ax + b$ such that\n$$15x^4 + px^3 - 37x^2 + qx + 6 = (15x^3 + 26x^2 - 11x - 6)(ax + b).$$\n\n3. **Multiply the polynomials:**\n$$(15x^3 + 26x^2 - 11x - 6)(ax + b) = 15a x^4 + 15b x^3 + 26a x^3 + 26b x^2 - 11a x^2 - 11b x - 6a x - 6b.$$\nGroup like terms:\n$$= 15a x^4 + (15b + 26a) x^3 + (26b - 11a) x^2 + (-11b - 6a) x - 6b.$$\n\n4. **Match coefficients with the quartic polynomial:**\nEquate coefficients of corresponding powers of $x$:\n- Coefficient of $x^4$: $15a = 15$\n- Coefficient of $x^3$: $15b + 26a = p$\n- Coefficient of $x^2$: $26b - 11a = -37$\n- Coefficient of $x$: $-11b - 6a = q$\n- Constant term: $-6b = 6$\n\n5. **Solve for $a$ and $b$ first:**\nFrom $15a = 15$, divide both sides by 15:\n$$15\cancel{a} = 15 \Rightarrow \cancel{15}a = \cancel{15} \Rightarrow a = 1.$$\nFrom $-6b = 6$, divide both sides by $-6$:\n$$\cancel{-6}b = 6 \Rightarrow b = \frac{6}{-6} = -1.$$\n\n6. **Find $p$ and $q$ using $a=1$ and $b=-1$:**\nCalculate $p$:\n$$p = 15b + 26a = 15(-1) + 26(1) = -15 + 26 = 11.$$\nCalculate $q$:\n$$q = -11b - 6a = -11(-1) - 6(1) = 11 - 6 = 5.$$\n\n**Final answer:**\n$$p = 11, \quad q = 5.$$