which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?\no f(x)=(x…

Explanation:

Step1: Recall complex - conjugate root theorem

If a polynomial with real - valued coefficients has a complex root (a + bi), then its complex conjugate (a - bi) is also a root. Given roots (2i) and (3i), the conjugate roots are (- 2i) and (-3i).

Step2: Write the polynomial in factored form

A polynomial (f(x)) with roots (r_1,r_2,\cdots,r_n) can be written as (f(x)=a(x - r_1)(x - r_2)\cdots(x - r_n)), where (a) is the leading coefficient. Since (a = 1) and the roots are (2i,-2i,3i,-3i), we have (f(x)=(x + 2i)(x - 2i)(x+3i)(x - 3i)).

Answer:

D. (f(x)=(x + 2i)(x + 3i)(x - 2i)(x - 3i))