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

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

c

step-by-step explanation:

c