$$(-x-9)(x^2+12x+27) = 0$$

$$ \begin{matrix}x_1 = -3 & x_2 = -9 \\[1 em] \end{matrix} $$

$ \color{blue}{ -x^{3}-21x^{2}-135x-243 } $ 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 (-243) are 1 3 9 27 81 243 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 243 }{ 1 } ~ $$

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(-3) = 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 + 3} $

$$ \frac{ -x^{3}-21x^{2}-135x-243 }{ \color{blue}{ x + 3 } } = -x^{2}-18x-81 $$

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.

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