$$\frac{x^5+33x^4+430x^3+2750x^2+8529x+10017}{x+3} = 0$$
Answer
Explanation
$ \color{blue}{ x^{5}+33x^{4}+430x^{3}+2750x^{2}+8529x+10017 } $ is a polynomial of degree 5. 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 (10017) are 1 3 7 9 21 27 53 63 159 189 371 477 1113 1431 3339 10017 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 53 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 159 }{ 1 } , ~ \pm \frac{ 189 }{ 1 } , ~ \pm \frac{ 371 }{ 1 } , ~ \pm \frac{ 477 }{ 1 } , ~ \pm \frac{ 1113 }{ 1 } , ~ \pm \frac{ 1431 }{ 1 } , ~ \pm \frac{ 3339 }{ 1 } , ~ \pm \frac{ 10017 }{ 1 } ~ $$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} $
$$ \frac{ x^{5}+33x^{4}+430x^{3}+2750x^{2}+8529x+10017 }{ \color{blue}{ x + 3 } } = x^{4}+30x^{3}+340x^{2}+1730x+3339 $$Polynomial $ x^{4}+30x^{3}+340x^{2}+1730x+3339 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{4}+30x^{3}+340x^{2}+1730x+3339 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.