# Thread: Proof of a polynomial function having a solution in R.

1. ## Proof of a polynomial function having a solution in R.

Don't really know what I'm doing with this proof here a little help would be amazing right now. Here is the question.

Let
f(x)=a0+a1x+...+anx^n (x ϵ ℝ)
be a polynomial function, an ≠ 0. Suppose n is odd. show that the equation f(x)=0 has at least one solution in ℝ.

2. Originally Posted by ml692787
Don't really know what I'm doing with this proof here a little help would be amazing right now. Here is the question.

Let
f(x)=a0+a1x+...+anx^n (x ϵ ℝ)
be a polynomial function, an ≠ 0. Suppose n is odd. show that the equation f(x)=0 has at least one solution in ℝ.
A proof can go along these lines.

Without loss of generality, assume $\displaystyle a_n > 0$

since n is odd, and $\displaystyle f(x)$ is continuous, we have $\displaystyle \lim_{x \to \infty}f(x) > 0$ and $\displaystyle \lim_{x \to - \infty} f(x) < 0$

thus, by the intermediate value theorem, there exists a real number $\displaystyle x \in (-\infty, \infty)$ such that $\displaystyle f(x) = 0$

3. Thanks a lot this makes a lot of sense after reading the chapter and putting it together with your thoughts, very much appreciated!

4. I like to do it with sequences instead. It is more athestically pleasing. (However, the proof using field theory is even more nicer but this is not an algebra thread.)

Let $\displaystyle f(x)$ be a polynomial in $\displaystyle \mathbb{R}$ of odd degree. Define $\displaystyle x_n = f(n)$. This is a sequence. Now argue that $\displaystyle \lim \ x_n = + \infty$ without lose of generality if the leading coefficient it positive. And define $\displaystyle y_n = f(-n)$ and show $\displaystyle \lim \ y_n = -\infty$. Thus, this means by definition there is a number $\displaystyle n_1<0$ so that $\displaystyle f(n_1)<0$ and there is a number $\displaystyle n_2>0$ so that $\displaystyle f(n_2)>0$. Thus, there is a zero by intermediate value theorem. Q.E.D.