$$s^5+2s^4+24s^3+48s^2-25s-50 = 0$$

$$ \begin{matrix}s_1 = 1 & s_2 = -1 & s_3 = -2 \\[1 em] s_4 = 5 i & s_5 = -5 i \\[1 em] \end{matrix} $$

$ \color{blue}{ x^{5}+2x^{4}+24x^{3}+48x^{2}-25x-50 } $ 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 (-50) are 1 2 5 10 25 50 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } ~ $$

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} $

$$ \frac{ x^{5}+2x^{4}+24x^{3}+48x^{2}-25x-50 }{ \color{blue}{ x - 1 } } = x^{4}+3x^{3}+27x^{2}+75x+50 $$

Polynomial $ x^{4}+3x^{3}+27x^{2}+75x+50 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{4}+3x^{3}+27x^{2}+75x+50 $

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

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