If I have a polynomial of degree $n$ with leading coefficient $1$, that when set equal to zero has as its only solution $x=0$, how do I prove that this polynomial can only be $x^n$?
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1Do you mean "real solutions"? If yes, you do not, since $x^3+x$. – Alexey Burdin Apr 30 '15 at 23:44
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Over the complex numbers you know that a polynomial of degree $n$ can be written as $c \prod_{i=1}^n (x- r_i)$ with $c,r_i$ complex numbers.
Since the polynomial has leading coefficient $1$ the $c$ is $1$. Moreover, each $r_i$ is a zero/root of the polynomial.
So all $r_i$ must equal $0$ and you are done.
(This only works for the complex numbers, or more generally an algebraically closed field, yet not over the reals or rationals.)
quid
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Write $\,f(x) = x^n g(x),\ g(0)\neq 0.\,$ If $\,\deg g > 0\,$ then $\,g(c)=0\,$ for some $\,0\neq c\in \Bbb C\,$ so $\,f(c) = 0,\,$ contra hypothesis. Thus $\,\deg g = 0\,$ so $\,g = 1,\,$ by lead coef of $f = 1.$
Bill Dubuque
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