Since the divisor polynomial of degree 2 (quadratic), we add a polynomial of degree 1. M x + K.

P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)

Q(x) = x² + 2x - 3

- 3 has factors 3 and -1 whose sum is 2, coefficient of x.

Q(x) = (x + 3) ( x - 1)

If P(x) is to be exactly divisible by Q(x) then P(x) must be exactly divisible without a reminder by x + 3 as well as x - 1. Then P(-3) and P(1) must both be zero.

P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)

P(1) = 1 + 2 - 2 + 1 - 1 + M + K = 0

M + K = -1 ---- (1)

P(-3) = 81 - 54 - 18 - 3 - 1 - 3 M + K = 0

2 M - K = 5 --- (2)

Solve the two equations: adding them: 3 M = 4

M = 4/3

K = - 7/3

Hence, we can add: 4/2 x - 7/3 to the given polynomial to have the given polynomial exactly divisible by the divisor polynomial.