$$x^4+16x^3+47x^2-100x-132 = 0$$
$$ \begin{matrix}x_1 = -1 & x_2 = 2 & x_3 = -6 \\[1 em] x_4 = -11 & \\[1 em] \end{matrix} $$
$ \color{blue}{ x^{4}+16x^{3}+47x^{2}-100x-132 } $ is a polynomial of degree 4. 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 (-132) are 1 2 3 4 6 11 12 22 33 44 66 132 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 11 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 22 }{ 1 } , ~ \pm \frac{ 33 }{ 1 } , ~ \pm \frac{ 44 }{ 1 } , ~ \pm \frac{ 66 }{ 1 } , ~ \pm \frac{ 132 }{ 1 } ~ $$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} $
$$ \frac{ x^{4}+16x^{3}+47x^{2}-100x-132 }{ \color{blue}{ x + 1 } } = x^{3}+15x^{2}+32x-132 $$Polynomial $ x^{3}+15x^{2}+32x-132 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+15x^{2}+32x-132 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.