$$16x^5-8x^3+x = 0$$
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 2 } & x_3 = -\dfrac{ 1 }{ 2 } \end{matrix} $$
In order to solve $ \color{blue}{ 16x^{5}-8x^{3}+x = 0 } $, first we need to factor our $ x $.
$$ 16x^{5}-8x^{3}+x = x \left( 16x^{4}-8x^{2}+1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 16x^{4}-8x^{2}+1 = 0$.
$ \color{blue}{ 16x^{4}-8x^{2}+1 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.
The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.
The factors of the leading coefficient ( 16 ) are 1 2 4 8 16 .The factor of the constant term ( 1 ) is 1 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 16 } ~ $$Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
If we plug these values into the polynomial $ P(x) $, we obtain $ P(\frac{ 1 }{ 2 }) = 0 $.
To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:
Divide $ P(x) $ with $ \color{blue}{ 2 x - 1 } $
$$ \frac{ 16x^{4}-8x^{2}+1 }{ \color{blue}{ 2x - 1 } } = 8x^{3}+4x^{2}-2x-1 $$Polynomial $ 8x^{3}+4x^{2}-2x-1 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 8x^{3}+4x^{2}-2x-1 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.