Find all real solutions of the equation $x^{10} - x^8 + 8x^6 - 24x^4 + 32x^2 - 48 = 0$ I have been able to factorize the polynomial as follows: $$(x^2 - 2)(x^8 + x^6 + 10x^4 - 4x^2 + 24)$$ from which $\sqrt2$ and $-\sqrt2$ are obvious solutions. My guess is that $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ does not have any real solutions, but how would I show that?

Try to write $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ as sum of squares and positive constant.
There is only one negative term, and else are all square already. So I try to factorize the negative term into a square, like
$x^8 + x^6 + 9x^4 + x^4 - 4x^2 + 4 + 20 =...>0$
You can try to finish the rest.
