$$x^3+11x^2+30x-360 = 0$$
Answer
Explanation
$ \color{blue}{ x^{3}+11x^{2}+30x-360 } $ 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 (-360) are 1 2 3 4 5 6 8 9 10 12 15 18 20 24 30 36 40 45 60 72 90 120 180 360 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 120 }{ 1 } , ~ \pm \frac{ 180 }{ 1 } , ~ \pm \frac{ 360 }{ 1 } ~ $$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(4) = 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 - 4} $
$$ \frac{ x^{3}+11x^{2}+30x-360 }{ \color{blue}{ x - 4 } } = x^{2}+15x+90 $$Polynomial $ x^{2}+15x+90 $ can be used to find the remaining roots.
$ \color{blue}{ x^{2}+15x+90 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.