(x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n) then

We have,

(x³ + 10x²mx + n)

Let, p (x) = x³ + 10x² + mx + n

It is given in the question that (x + 2) and (x – 1) are the factors of p (x)

∴ p (-2) = 0

(-2)³ + 10 (-2)² + m (-2) + n = 0

- 8 + 40 – 2m + n = 0

32 – 2m + n = 0

2m – n – 32 = 0 ……….(i)

And,

p (1) = 0

(1)³+ 10 (1)² + m (1) n

1 + 10 + m + n = 0

11 + m + n = 0

m + n + 11 = 0 ………. (ii)

Now, adding equation (i) and (ii) we get

2m – n - 32 + m + n + 11 = 0

3m - 21 = 0

3m = 21

m =

m = 7

Now, putting the value of m in (ii) we get:

7 + n + 11 = 0

18 + n = 0

n = - 18

∴ the value of m is 7 and that of n is – 18

Hence, option B is correct