$ \color{blue}{ 625x^{4}-450x^{2}+81 } $ 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 ( 625 ) are 1 5 25 125 625 .The factors of the constant term (81) are 1 3 9 27 81 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 1 }{ 625 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 25 } , ~ \pm \frac{ 3 }{ 125 } , ~ \pm \frac{ 3 }{ 625 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 5 } , ~ \pm \frac{ 9 }{ 25 } , ~ \pm \frac{ 9 }{ 125 } , ~ \pm \frac{ 9 }{ 625 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 5 } , ~ \pm \frac{ 27 }{ 25 } , ~ \pm \frac{ 27 }{ 125 } , ~ \pm \frac{ 27 }{ 625 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 81 }{ 5 } , ~ \pm \frac{ 81 }{ 25 } , ~ \pm \frac{ 81 }{ 125 } , ~ \pm \frac{ 81 }{ 625 } ~ $$

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{ 3 }{ 5 }) = 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}{ 5 x - 3 } $

$$ \frac{ 625x^{4}-450x^{2}+81 }{ \color{blue}{ 5x - 3 } } = 125x^{3}+75x^{2}-45x-27 $$

Polynomial $ 125x^{3}+75x^{2}-45x-27 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 125x^{3}+75x^{2}-45x-27 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

Polynomial Equations Solver