Find the polynomial with real coefficients, smallest possible degree, leading coefficient 1, such that 2 and 3+i are zeroes of the polynomial.

f (x) = x³ - 8x² + 22x - 20

given x = a, x = b are zeros of a polynomial then

(x - a) and (x - b) are factors and the polynomial is the product of the factors

f (x) = k (x - a) (x - b) → (k is a multiplier)

note that complex zeros occur in conjugate pairs

3 + 1 is a zero then 3 - i is a zero

zeros are x = 2, x = 3 + i and x = 3 - i, thus

(x - 2), (x - (3 + i)) and (x - (3 - i)) are the factors

f (x) = (x - 2) (x - 3 - i) (x - 3 + i)

= (x - 2) (x² - 6x + 10)

= x³ - 8x² + 22x - 20