Write a polynomial function in standard form having the roots of -1, 2 and 3i. Use a

leading coefficient of 1.

P(x) = x³-(3i+1)x²+(3i-2)x+6i

Step-by-step explanation:

If -1, 2 and 3i are roots of the polynomial, then the following are factors of the polynomial in x

(x+1), x-2 and x-3i

To get the required polynomial, we will take the product of the factors as shown;

(x+1)(x-2)(x-3i)

(x+1)(x-2) = x²-2x+x-2

(x+1)(x-2) = x²-x-2

(x²-x-2)(x-3i) = (x²)(x)-x²(3i)-x(x)-x(-3i)-2x-2(-3i)

(x²-x-2)(x-3i) = x³-3x²i-x²+3xi-2x+6i

(x²-x-2)(x-3i) = x³-x²(3i+1)+x(3i-2)+6i

(x²-x-2)(x-3i) = x³-(3i+1)x²+(3i-2)x+6i

Hence the polynomial function in standard form with leading coefficient of 1 is x³-(3i+1)x²+(3i-2)x+6i

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

i think the answer is d im not sure