Using the remainder theorem, find the remainders obtained when x³ + (kx + 8)x + k is divided by x + 1 and x - 2. Hence, find k if the sum of two remainders is 1.

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = x³ + (kx + 8)x + k by x + 1 or (x - (-1))

Remainder = r₁ = f(-1) = -1³ + ((-1)k + 8)(-1) + k
= -1 + (8 - k)(-1) + k
= -1 - 8 + k + k
= 2k - 9

∴ On dividing, f(x) = x³ + (kx + 8)x + k by x - 2

Remainder = r₂ = (2)³ + (k(2) + 8)2 + k
= 8 + 4k + 16 + k
= 5k + 24

Given, sum of two remainders = 1

∴ r₁ + r₂ = 1

$\Rightarrow 2k - 9 + 5k + 24 = 1 \\[0.5em] \Rightarrow 7k + 15 = 1 \\[0.5em] \Rightarrow 7k = 1 - 15 \\[0.5em] \Rightarrow 7k = -14 \\[0.5em] \Rightarrow k = -\dfrac{14}{7} \\[0.5em] k = -2.$

The first remainder is 2k - 9 and the second remainder is 5k + 24 and the value of k is -2.