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

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: