$$s(s+7)(s+9)(s+10)(s+15) = 0$$

$$ \begin{matrix}s_1 = 0 & s_2 = -7 & s_3 = -9 \\[1 em] s_4 = -10 & s_5 = -15 \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ x^{5}+41x^{4}+613x^{3}+3975x^{2}+9450x = 0 } $, first we need to factor our $ x $.

$$ x^{5}+41x^{4}+613x^{3}+3975x^{2}+9450x = x \left( x^{4}+41x^{3}+613x^{2}+3975x+9450 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ x^{4}+41x^{3}+613x^{2}+3975x+9450 = 0$.

$ \color{blue}{ x^{4}+41x^{3}+613x^{2}+3975x+9450 } $ 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 (9450) are 1 2 3 5 6 7 9 10 14 15 18 21 25 27 30 35 42 45 50 54 63 70 75 90 105 126 135 150 175 189 210 225 270 315 350 378 450 525 630 675 945 1050 1350 1575 1890 3150 4725 9450 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 35 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 70 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 105 }{ 1 } , ~ \pm \frac{ 126 }{ 1 } , ~ \pm \frac{ 135 }{ 1 } , ~ \pm \frac{ 150 }{ 1 } , ~ \pm \frac{ 175 }{ 1 } , ~ \pm \frac{ 189 }{ 1 } , ~ \pm \frac{ 210 }{ 1 } , ~ \pm \frac{ 225 }{ 1 } , ~ \pm \frac{ 270 }{ 1 } , ~ \pm \frac{ 315 }{ 1 } , ~ \pm \frac{ 350 }{ 1 } , ~ \pm \frac{ 378 }{ 1 } , ~ \pm \frac{ 450 }{ 1 } , ~ \pm \frac{ 525 }{ 1 } , ~ \pm \frac{ 630 }{ 1 } , ~ \pm \frac{ 675 }{ 1 } , ~ \pm \frac{ 945 }{ 1 } , ~ \pm \frac{ 1050 }{ 1 } , ~ \pm \frac{ 1350 }{ 1 } , ~ \pm \frac{ 1575 }{ 1 } , ~ \pm \frac{ 1890 }{ 1 } , ~ \pm \frac{ 3150 }{ 1 } , ~ \pm \frac{ 4725 }{ 1 } , ~ \pm \frac{ 9450 }{ 1 } ~ $$

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(-7) = 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 + 7} $

$$ \frac{ x^{4}+41x^{3}+613x^{2}+3975x+9450 }{ \color{blue}{ x + 7 } } = x^{3}+34x^{2}+375x+1350 $$

Polynomial $ x^{3}+34x^{2}+375x+1350 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}+34x^{2}+375x+1350 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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