Find roots of polynomial $$ p(x) = -29s^3-35s^2-2s^5-17s^4-45s $$

The roots of polynomial $ p(s) $ are:

$$ \begin{aligned}s_1 &= 0\\[1 em]s_2 &= -1.7545\\[1 em]s_3 &= -6.6351\\[1 em]s_4 &= -0.0552+1.3892i\\[1 em]s_5 &= -0.0552-1.3892i \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} -29s^3-35s^2-2s^5-17s^4-45s & = 0\\[1 em] -2s^5-17s^4-29s^3-35s^2-45s & = 0 \end{aligned} $$

Step 2:

Factor out $ \color{blue}{ -s }$ from $ -2s^5-17s^4-29s^3-35s^2-45s $ and solve two separate equations:

$$ \begin{aligned} -2s^5-17s^4-29s^3-35s^2-45s & = 0\\[1 em] \color{blue}{ -s }\cdot ( 2s^4+17s^3+29s^2+35s+45 ) & = 0 \\[1 em] \color{blue}{ -s = 0} ~~ \text{or} ~~ 2s^4+17s^3+29s^2+35s+45 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ s = 0 } $. Use second equation to find the remaining roots.

Step 3:

Polynomial $ 2s^4+17s^3+29s^2+35s+45 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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