If (2x + 5y)² – 5(2x + 5y) – 14 = (2x + 5y + p)(2x + 5y + q), then the value…

Concept

When a compound expression such as (2x + 5y) repeats throughout an equation, substituting a single variable for it converts the equation into a standard quadratic. Comparing the coefficients of a quadratic u2 + bu + c with its factorised form (u + p)(u + q) = u2 + (p + q)u + pq lets the coefficient of u be read off directly as p + q, without solving for p and q individually.

Step-by-step solution

1. Let u = 2x + 5y. The given equation becomes u2 − 5u − 14 = (u + p)(u + q).

2. Expand the right-hand side: (u + p)(u + q) = u2 + (p + q)u + pq.

3. Compare the coefficient of u on both sides: the coefficient of u on the left is −5, so p + q = −5.

Cross-check

Solving u2 − 5u − 14 = 0 by the quadratic formula gives u = [5 ± √(25 + 56)] / 2 = [5 ± 9] / 2, i.e. u = 7 or u = −2. Since u2 − 5u − 14 = (u − 7)(u + 2), matching this with (u + p)(u + q) gives p + q = −7 + 2 = −5, the same value obtained from direct coefficient comparison.

Hence, p + q = −5.