$$(x+4)^2(x-3)^3 = 0$$
Answer
Explanation
$ \color{blue}{ x^{5}-x^{4}-29x^{3}+45x^{2}+216x-432 } $ 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 (-432) are 1 2 3 4 6 8 9 12 16 18 24 27 36 48 54 72 108 144 216 432 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 144 }{ 1 } , ~ \pm \frac{ 216 }{ 1 } , ~ \pm \frac{ 432 }{ 1 } ~ $$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} $
$$ \frac{ x^{5}-x^{4}-29x^{3}+45x^{2}+216x-432 }{ \color{blue}{ x - 3 } } = x^{4}+2x^{3}-23x^{2}-24x+144 $$Polynomial $ x^{4}+2x^{3}-23x^{2}-24x+144 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{4}+2x^{3}-23x^{2}-24x+144 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.