Answer:
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
