Question

Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?

x) = (x – 2i)(x – 3i)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)

Answer 1

In the polynomial if the roots are complex numbers then they are always exists in pair.

Given roots are x= 2i, x = 3i so other roots are x= - 2i and x = -3i

So the factors of the polynomials are (x-2i)(x+2i)(x-3i)(x+3i)

As the multiplicity of the roots is 1 so the exponent of each parenthesis is 1.

So the polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1 =  (x-2i)(x+2i)(x-3i)(x+3i)

So the option 4 is the  answer (x + 2i)(x + 3i)(x – 2i)(x – 3i)

Answer 2

Answer:

D)   f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)

Step-by-step explanation: