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

[Muffins]

Homework Statement

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)

Homework Equations

(a+bi)*(a-bi) = a^2 + b^2 where a, and b are always reals?
Not really sure if this helps or not.

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!

 

[Dick]

The root is where f(c)=0. Take the complex conjugate of that equation.

 

[Muffins]

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.

 

[Muffins]

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