Complex zeros of polynomials.

Question

Write a polynomial of degree two in x with REAL coefficients and leading coefficient 1 for which 4-3i is a root. 

Answer 1

Using the quadratic formula we can find the values for a, b, and c and then find the polynomial

 

4 -3i = ( b - sqrt(b² -4ac))/2a

 

we know a = 1   (a is the leading coefficient)

 

so 2a = 2 and we will multiply both sides by 2

 

8 -6i = b - sqrt(b² -4ac)

 

from this expression we can break it up into it's real and imaginary pieces

 

so b = 8

 

and 6i = sqrt(b2 -4ac)

 

so

 

6i = sqrt(64 -4ac)

 

square both sides

 

-36 = 64 -4ac  (a = 1 from above)

 

-36 = 64 - 4c

 

-100 = -4c

 

c =25

 

our equation would be

 

ax² +bx +c =0

 

now put in a, b and c that we found

 

x² +8x +25 = 0

 

Hope this helps