write a polynomial function of f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros, 3i, 2, -i

Question

I understand the basic concepts of the question and know that i= the square root of -1 but I'm not sure how to incorporate the i

Answer 1

Complex roots come in conjugates.  Your roots are

 

x = -3i

x= 3i

x = 2

x = -i

x = i

 

Since  i=√-1 , then i²=-1

 

For the root   x = 3i , we can say that the factor is  

 

(x - 3i)(x + 3i) = x² - 9i²

                      = x² - 9(-1)

                      = x² + 9

 

(x - i)(x + i) = x² - i²

                  = x² + 1

 

 

Then your polynomial is

 

f(x) = (x - 2)(x² + 1)(x² + 9)

Answer 2

If the coefficients are real, and some zeros are complex, then the complex conjugate of each complex zero must also be a zero.  Therefore, -3i and i are also zeros.  That makes 5 zeros, so the minimum degree is 5.  With a leading coefficient of 1, we get

 

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

 

Note that

 

(x - 3i)(x + 3i) = x² + 9

(x + i)(x - i) = x² + 1

 

Then

 

f(x) = (x² + 9)(x - 2)(x² + 1)

 

Note the coefficients are all real, in spite of the 4 complex zeros.  You may expand f(x) into standard form if you wish.