$$(5x+9)^3 = 0$$
Answer
Explanation
$ \color{blue}{ 125x^{3}+675x^{2}+1215x+729 } $ 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 ( 125 ) are 1 5 25 125 .The factors of the constant term (729) are 1 3 9 27 81 243 729 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 25 } , ~ \pm \frac{ 3 }{ 125 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 5 } , ~ \pm \frac{ 9 }{ 25 } , ~ \pm \frac{ 9 }{ 125 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 5 } , ~ \pm \frac{ 27 }{ 25 } , ~ \pm \frac{ 27 }{ 125 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 81 }{ 5 } , ~ \pm \frac{ 81 }{ 25 } , ~ \pm \frac{ 81 }{ 125 } , ~ \pm \frac{ 243 }{ 1 } , ~ \pm \frac{ 243 }{ 5 } , ~ \pm \frac{ 243 }{ 25 } , ~ \pm \frac{ 243 }{ 125 } , ~ \pm \frac{ 729 }{ 1 } , ~ \pm \frac{ 729 }{ 5 } , ~ \pm \frac{ 729 }{ 25 } , ~ \pm \frac{ 729 }{ 125 } ~ $$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{ 9 }{ 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 + 9 } $
$$ \frac{ 125x^{3}+675x^{2}+1215x+729 }{ \color{blue}{ 5x + 9 } } = 25x^{2}+90x+81 $$Polynomial $ 25x^{2}+90x+81 $ can be used to find the remaining roots.
$ \color{blue}{ 25x^{2}+90x+81 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.