<s>[INST] Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$ Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$
I entered the equation in desmos.com and no roots were lying below or x=0 lines , hence all roots are imaginary.
Using Descartes rule for f(x) 4 sign change occurs , hence 4,2 or zero positive roots.
Using Descartes rule for f(-x) 0 sign change occurs , hence zero negative roots.
For here on how do I proceed
 [/INST] To expand on Dr. Sonnhard Graubner's answer, let $f(x) = 54x^4-36x^2+15x^2-6x+1$. Then:
$$54x^4-36x^3+18x^2-6x+1 > f(x)$$
$$\Rightarrow (3x-1)^2(6x^2+1)$$
by the rational root test. 
The smaller function has only one root: $x = \frac{1}{3}$, and there $f''(x) = 648x^2-216x+30$ is positive. Therefore, for all real $x$, $f(x) > 0$ and therefore $54x^4-36x^3+18x^2-6x+1 > 0$.
</s>