In polynomial algebra, consider the cubic polynomial p·x^3 − q·x^2 − 7x − 6. If this polynomial is exactly divisible by x^2 − x − 6 (that is, its remainder is zero), determine the values of the constants p and q.

Introduction / Context:
This question uses the idea of polynomial divisibility. If a cubic polynomial is divisible by a quadratic factor, then the roots of the quadratic must also be roots of the cubic. This allows us to set up equations involving the unknown parameters p and q by substituting the roots into the cubic and setting the resulting values to zero. This is a standard technique in algebra for determining unknown coefficients in polynomials.

Given Data / Assumptions:
- The cubic polynomial is p·x^3 − q·x^2 − 7x − 6.
- It is divisible by x^2 − x − 6.
- The quadratic factor x^2 − x − 6 can be factored further, which reveals its roots.
- We need to find exact values of p and q such that the divisibility condition holds.

Concept / Approach:
First, factor the quadratic x^2 − x − 6 to find its roots. If x^2 − x − 6 = (x − 3)(x + 2), then the roots are x = 3 and x = −2. If the cubic is divisible by this quadratic, then x = 3 and x = −2 must also be roots of the cubic p·x^3 − q·x^2 − 7x − 6. Therefore, substituting x = 3 and x = −2 into the cubic must give zero, yielding two linear equations in p and q that can be solved simultaneously.

Step-by-Step Solution:
1) Factor the quadratic: x^2 − x − 6 = (x − 3)(x + 2), so the roots are 3 and −2. 2) Let f(x) = p·x^3 − q·x^2 − 7x − 6. 3) For x = 3, divisibility implies f(3) = 0. Compute f(3) = p·3^3 − q·3^2 − 7·3 − 6 = 27p − 9q − 21 − 6. 4) Simplify: f(3) = 27p − 9q − 27. Set this equal to zero to get 27p − 9q − 27 = 0, or dividing by 9, 3p − q − 3 = 0. 5) For x = −2, we require f(−2) = 0. Compute f(−2) = p·(−8) − q·4 − 7·(−2) − 6 = −8p − 4q + 14 − 6. 6) Simplify: f(−2) = −8p − 4q + 8. Set this equal to zero: −8p − 4q + 8 = 0, or dividing by −4, 2p + q − 2 = 0. 7) Solve the system 3p − q − 3 = 0 and 2p + q − 2 = 0. Adding them gives 5p − 1 = 0, so p = 1. Then 2(1) + q − 2 = 0 gives q = 0.

Verification / Alternative check:
Substitute p = 1 and q = 0 back into the cubic, giving x^3 − 7x − 6. Factor the quadratic divisor x^2 − x − 6. Perform polynomial long division or factor x^3 − 7x − 6 as (x^2 − x − 6)(x + 1) to check divisibility. Multiplying (x^2 − x − 6)(x + 1) gives x^3 − 7x − 6, confirming that with p = 1 and q = 0 the cubic is indeed divisible by the quadratic with no remainder.

Why Other Options Are Wrong:
Other option pairs for p and q fail the divisibility test. For example, choosing p = 0 or q = 1 leads to values of f(3) or f(−2) that are non zero, meaning at least one of the quadratic roots is not a root of the cubic. Only the pair p = 1, q = 0 satisfies f(3) = 0 and f(−2) = 0 simultaneously, ensuring exact divisibility by x^2 − x − 6.

Common Pitfalls:
Mistakes often arise from incorrect substitution of the roots into the cubic or from arithmetic errors in simplifying coefficients. When solving the system for p and q, forgetting to divide by common factors can make the equations look more complicated than they are. It is also easy to assume divisibility from only one root and not check the other, which can lead to an incorrect conclusion. Systematic substitution and careful algebra avoid these traps.

Final Answer:
The only values that make the cubic polynomial p·x^3 − q·x^2 − 7x − 6 divisible by x^2 − x − 6 are p = 1, q = 0.