What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?
f(x) = x4 – 3x2 – 4
f(x) = x3 + x2 – 4x – 4

which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?
f(x) = –x2 – 8x – 16
f(x) = –x2 + 8x – 16
f(x) = –x2 – 8x + 16

which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1
f(x) = (x + 2i)(x + 3i)
f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)

In the question "What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?" The given roots are i, -2 and 2. Recall that for any polynomial having complex root, the conjugate of the complex root is also a root of the polynomial, thus -i is also a root of the required equation. Thus the required equation is obtainrd thus: f(x) = (x - i)(x + i)(x - 2)(x + 2) = (x^2 + 1)(x^2 - 4) = x^4 - 4x^2 + x^2 - 4 = x^4 - 3x^2 - 4 In the question "Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?" The required equation is obtained thus: f(x) = -(x - 4)^2 = -(x^2 - 8x + 16) = -x^2 + 8x - 16 In the question "Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1" Recall that for any polynomial having complex root, the conjugate of the complex root is also a root of the polynomial, thus -2i and -3i are also roots of the required equation. Thus the required equation is obtained thus: f(x) = (x + 2i)(x + 3i)(x - 2i)(x - 3i).

That’d be 9 years or something