# A polynomial with only one real $x$-intercept without imaginary roots has a root equal to $-c_n/(nc_{n-1})$.

Given coefficients $c_n, c_{n-1}, c_{n-2}, \ldots$ of the polynomial $c_n x^n+c_{n-1}x^{n-1} + \cdots +c_{1}x+c_0,$ prove that for $c_nx^n+c_{n-1}x^{n-1} + \cdots +c_1 x+c_0 = 0,$ $x=-c_n/(nc_{n-1})$.

For clarification, there are $n+1$ terms in the polynomial, which has only one real $x$-intercept and no imaginary roots.

• What does $-c_{n - 1}/c_n/n$ denote? – Travis Willse Nov 27 '15 at 18:19
• Which of the following does $-c_{n-1}/c_n/n$ denote? $$-\frac{c_{n-1}}{\frac{c_n}{n}}\\ -\frac{c_{n-1}}{n\cdot c_n}$$ – user228113 Nov 27 '15 at 18:22
• $-c_{n-1}/-c_{n}/n$ denotes the second. Sorry, for the confusion. – Max Li Nov 27 '15 at 19:48

If your polynomial has only one real root $r$ and no imaginary roots, then that means that the multiplicity of $r$ is $n$.
Therefore, we can write : $$c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 = c_n (x - r)^{n}$$ By expanding the RHS (using the binomial theorem), we find that the coefficient of $x^{n-1}$ is $- c_n \cdot n \cdot r$. Then, by identifying with the coefficient of $x^{n-1}$ in the LHS, we get $c_{n-1} = - c_n \cdot n \cdot r$.
So $r = -\dfrac{c_{n-1}}{n \cdot c_n}$.