Conjugate Zeros

Question

a. Find the quadratic polynomial with the zeros x=i and x=2i and leading coefficient of 1

 

b. This polynomial has no complex conjugate zeros. Why does this not contradict the complex zeros theorem?

Answer 1

Hi Naomi,

 

a) Let P(x) = ax^2 + bx + c We are given that a = 1 and we know the zeros. So, we can write

P(x) = (x - i)(x - 2i)

= x^2 - 3ix - 2

 

b) We can see that P(x) has complex coefficients, because b = 3i. By definition, Complex Conjugates Theorem only applies to polynomials with REAL coefficients so there is nothing to contradict. P(x) falls under the umbrella of the Fundamental Theorem of Algebra, which states that if f P(x) is a polynomial function of degree n (n > 0) with complex coefficients, then the equation P(x) = 0 has n roots assuming you count double roots as 2, triple roots as 3, etc.