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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

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2 Answers

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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 $54x^4-36x^3+18x^2-6x+1 > 0$.

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Hint: Prove that $$54x^4-36x^3+18x^2-6x+1>0$$ for all real $$x$$

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