Complex Conjugate Zeros
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Key Questions

A complex conjugate is the number to which you have to multiply a complex number in order to make it real.
By using the identity#(x+y) . (xy) = xÂ²yÂ²# , we see that, to every complex, there is another to which we can multiply it in order to get a new number that will not depend on#i# .
If#(a+bi).(c+di)# is real, (#c+di# ) is (#a+bi# )'s conjugate and it equals (#abi# ). 
The complex conjugate of a complex number
#a+bi# can be found by negating the imaginary part, that is,#abi# .For example, the complex conjugate of
#23i# is#2+3i# . 
Answer:
If a polynomial has Real coefficients, then any Complex zeros will occur in Complex conjugate pairs.
That is, if
#z = a+bi# is a zero then#bar(z) = abi# is also a zero.Explanation:
Actually a similar theorem holds for square roots and polynomials with rational coefficients:
If
#f(x)# is a polynomial with rational coefficients and a zero expressible in the form#a+b sqrt(c)# where#a, b, c# are rational and#sqrt(c)# is irrational, then#ab sqrt(c)# is also a zero. 
As a student, if a teacher tells you that a polynomial with real coefficients has
#3i# for one of its zeros, that you can reason:With real corrifients, if
#3i# is a zero, then so is#3i# . So I know that#(x3i)# and#(x(3i)) = (x+3i)# are both factors.So I know that
#(x3i)(x+3i) = x^2 + 9# is a factor. That might help me factor the polynomial. (If I've learned division of polynomials.)I saw this interesting bit of reasoning recently here on Socratic.
We know that a polynomial of degree
#n# has#n# zeros (counting multiplicity). So a polynomial of odd degree has an odd number of zeros.
We also know that if a polynomial has real coefficients, the any imaginary zeros occur in conjugate pairs.So we can conclude that a polynomial with real coefficients and odd degee must have at least one real zero. (An odd number of zeros cannot all be imaginary.)