$$x^4-8x^3+9x^2+18x = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -1 & x_3 = 3 \\[1 em] x_4 = 6 & \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ x^{4}-8x^{3}+9x^{2}+18x = 0 } $, first we need to factor our $ x $.

$$ x^{4}-8x^{3}+9x^{2}+18x = x \left( x^{3}-8x^{2}+9x+18 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ x^{3}-8x^{2}+9x+18 = 0$.

$ \color{blue}{ x^{3}-8x^{2}+9x+18 } $ 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 (18) are 1 2 3 6 9 18 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } ~ $$

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(-1) = 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 + 1} $

$$ \frac{ x^{3}-8x^{2}+9x+18 }{ \color{blue}{ x + 1 } } = x^{2}-9x+18 $$

Polynomial $ x^{2}-9x+18 $ can be used to find the remaining roots.

$ \color{blue}{ x^{2}-9x+18 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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