$f(x) = x^4 + bx^3 + 5x^2 + dx + 6$
By Remainder Theorem
When f(x) is divided by x - 2, the remainder is f(2)
$\color{#FC6255}{f(2) =} 2^4 + b(2^3) + 5(2^2) + d(2) + 6 = 16$
$8b + 2d = -26$ ← Equation (1)
When f(x) is divided by x + 1, the remainder is f(-1)
$\color{#FC6255}{f(-1) =} (-1)^4 + b(-1)^3 + 5(-1)^2 + d(-1) + 6 = 10$
$-b - d = -2$ ← Equation (2)
From Equation (1) and Equation (2)
$b = -5$
$d = 7$ ← answer
