Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 5, -3, -1+3i

Answer 1

The other zero must be the conjugate of -1+3i  which is -1-3i.

so in factor form the function is

(x-5)(x+3)(x - (-1+3i))(x -(-1-3i)

= (x-5)(x+3)( x^2 + x +3ix + x + 1 + 3i - 3ix - 3i -9i^2)

= (x-5)(x+3)(x^2 + 2x + 10)

= (x^2 - 2x - 15)(x^2 + 2x + 10)

=  x^4 - 9x^2 - 50x - 150

Answer 2

This is a polynomial of degree 4, because you have 4 roots real or imaginary:1st) x= 5 →(x-5)=0
2nd) x = - 3→(x+3) =0
3rd) x = -1+3i →(x+1-3i) = 0
and the 4th one that is not mentioned which is the conjugate of 
-1+3i, that is -1-3i (in any polynomial if a root has the form of a+bi, there is always a conjugate root = a-bi)
 
4th) x= -1-3i→(x+1+3i) = 0
Hence the polynomial =(x-5)(x+3)((x+1-3i)(x+1+3i)
Solving the above will give you (unless I am mistaken, pls recalculate):

x⁴-9x²-50x-150<u />