Any real root of a polynomial
A nice polynomial problem:
Find a real root of the equation $$(x^2 - 9x - 1)^{10} + 99x^{10} = 10x^9(x^2-1)$$
Looks scary with the tenth powers, but move $90x^9$ to the right to get
$$(x^2 - 9x - 1)^{10} + 9x^{10} = 10x^9(x^2 - 9x - 1)$$
Now get the terms to the left. A pattern begins to emerge.
$$(x^2 - 9x - 1)^{10} - 10x^9(x^2 - 9x - 1) + 9x^{10} = 0$$
Split the middle term with coefficients of $9$ and $1$:
$$(x^2 - 9x - 1)^{10} - 9x^9(x^2 - 9x - 1) - x^9(x^2 - 9x - 1) + 9x^{10} = 0$$
Factoring this is a bit tricky. We could do either of these:
$$ \begin{align} (x^2 - 9x - 1)[(x^2 - 9x - 1)^9 - 9x^9] - x^9[(x^2 - 9x - 1) - 9x] &= 0 \label{a}\tag{1} \\ (x^2 - 9x - 1)[(x^2 - 9x - 1)^9 - x^9] - 9x^9[(x^2 - 9x - 1) - x] &= 0 \label{b}\tag{2} \end{align} $$Note that $[(x^2 - 9x - 1)^9 - x^9] = (x^2 - 10x - 1)g(x)$, where $g(x)$ is some polynomial in x. Trying to do the same with $9x^9$ gives us the ninth root of nine, which gets messy. Let’s continue with $\ref{b}$.
$$(x^2 - 9x - 1)(x^2 - 10x - 1)g(x) - 9x^9(x^2 - 10x - 1) = 0$$
And we have a factor! Th roots of $x^2 - 10x - 1$ are $5 \pm \sqrt{26}$, giving us two real roots of the original equation.