Find the values of m and n so that x - 1 and x + 2 both are factors of

x³ + (3m + 1)x² + nx - 18.

x - 1 = 0 ⇒ x = 1.

Since, x - 1 is a factor of x³ + (3m + 1)x² + nx - 18,

∴ On substituting x = 1 in x³ + (3m + 1)x² + nx - 18, remainder = 0.

⇒ (1)³ + (3m + 1)(1)² + n(1) - 18 = 0

⇒ 1 + 3m + 1 + n - 18 = 0

⇒ 3m + n - 16 = 0

⇒ n = 16 - 3m ………(i)

x + 2 = 0 ⇒ x = -2.

Since, x + 2 is a factor of x³ + (3m + 1)x² + nx - 18,

∴ On substituting x = -2 in x³ + (3m + 1)x² + nx - 18, remainder = 0.

(-2)³ + (3m + 1)(-2)² + n(-2) - 18 = 0

⇒ -8 + (3m + 1)(4) - 2n - 18 = 0

⇒ -8 + 12m + 4 - 2n - 18 = 0

⇒ 12m - 2n - 22 = 0

⇒ 12m - 2n = 22

⇒ 2(6m - n) = 22

⇒ 6m - n = 11

⇒ n = 6m - 11 ………(ii)

From (i) and (ii) we get,

⇒ 16 - 3m = 6m - 11

⇒ 6m + 3m = 16 + 11

⇒ 9m = 27

⇒ m = 3.

Substituting m = 3 in (ii) we get,

⇒ n = 6(3) - 11 = 18 - 11 = 7.

Hence, m = 3 and n = 7.