# Show that complex conjugate is also a root of polynomial with real coefficients

by Muffins
Tags: coefficients, complex, conjugate, polynomial, real, root
 P: 6 1. The problem statement, all variables and given/known data Suppose that f(x) is a polynomial of degree n with real coefficients; that is, f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real) Suppose that c ∈ C(complex) is a root of f(x). Prove that c conjugate is also a root of f(x) 2. Relevant equations (a+bi)*(a-bi) = a^2 + b^2 where a, and b are always reals? Not really sure if this helps or not. 3. The attempt at a solution I'm really clueless on how to start approaching this. I was thinking perhaps the fundamental theorem of algebra might be of some use, or perhaps the fact that a number of complex form multiplied by it's conjugate is a real number, but I'm really not sure. Could anyone give me a nudge in the right direction? Any help would be greatly appreciated!! Thanks!
 Sci Advisor HW Helper Thanks P: 24,987 The root is where f(c)=0. Take the complex conjugate of that equation.
 P: 6 Never mind! I found the actual theorem on the web, and I think this is pretty much what I was looking for anyway Consider the polynomial f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real) where all a*x are real. The equation f(x) = 0 is thus a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n = 0 Given that all of the coefficients are real, we have a_n x^n(conjugate) = a_n x^n(x is conjugate) Thus it follows that a_n x^n(conj)+ a_(n-1) x^(n-1)(conj)+ …+a_1 x(conj)+ a_0, a_n = 0(conj) = 0 and thus that for any root ζ its complex conjugate is also a root.
P: 6

## Show that complex conjugate is also a root of polynomial with real coefficients

Hahah, thanks Dick! I caught on a little late, but thanks a bunch for your reply!

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