Question

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+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

The first option we have is the polynomial funtion:

$f\mleft(x\mright)=(x-2i)\mleft(x-3i\mright)$

To calculate the roots of this polynomial, we need to equal each parenthesis to zero, and solve for x:

$\begin{gathered} x-2i=0 \\ x=2i \end{gathered}$

And for the second parenthesis equal to zero we have:

$\begin{gathered} x-3i=0 \\ x=3i \end{gathered}$

This polynomial function has roots 2i and 3i, whith a multiplicity if 1 because they only appear 1 time as a root each.

Also, we need to find if the leading coefficient is 1, for this we need to multiply the two parentesis of the function:

$\begin{gathered} f\mleft(x\mright)=(x-2i)\mleft(x-3i\mright) \\ f(x)=x^2-3ix-2ix+6i^2 \end{gathered}$

The leading coefficient is the one with the x that has the highest power. Since the highest power is x^2 we can see that indeed it has a leading coefficient of 1.

Thus, the answer is the first option: f(x)= (x-2i)(x-3i)