$$x^5-24x^4+189x^3-486x^2 = 0$$
Answer
Explanation
In order to solve $ \color{blue}{ x^{5}-24x^{4}+189x^{3}-486x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ x^{5}-24x^{4}+189x^{3}-486x^{2} = x^2 \left( x^{3}-24x^{2}+189x-486 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ x^{3}-24x^{2}+189x-486 = 0$.
$ \color{blue}{ x^{3}-24x^{2}+189x-486 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (-486) are 1 2 3 6 9 18 27 54 81 162 243 486 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 162 }{ 1 } , ~ \pm \frac{ 243 }{ 1 } , ~ \pm \frac{ 486 }{ 1 } ~ $$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(6) = 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}{x - 6} $
$$ \frac{ x^{3}-24x^{2}+189x-486 }{ \color{blue}{ x - 6 } } = x^{2}-18x+81 $$Polynomial $ x^{2}-18x+81 $ can be used to find the remaining roots.
$ \color{blue}{ x^{2}-18x+81 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.