$$-2^3x^4+2\cdot2^3x^3-(2^3+2^2)x^2+(2^2-1)x = 0$$
Answer
Explanation
In order to solve $ \color{blue}{ -8x^{4}+16x^{3}-12x^{2}+3x = 0 } $, first we need to factor our $ x $.
$$ -8x^{4}+16x^{3}-12x^{2}+3x = x \left( -8x^{3}+16x^{2}-12x+3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -8x^{3}+16x^{2}-12x+3 = 0$.
$ \color{blue}{ -8x^{3}+16x^{2}-12x+3 } $ is a polynomial of degree 3. 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 ( -8 ) are 1 2 4 8 .The factors of the constant term (3) are 1 3 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 8 } ~ $$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 } $
$$ \frac{ -8x^{3}+16x^{2}-12x+3 }{ \color{blue}{ 2x - 1 } } = -4x^{2}+6x-3 $$Polynomial $ -4x^{2}+6x-3 $ can be used to find the remaining roots.
$ \color{blue}{ -4x^{2}+6x-3 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.