Find roots of polynomial $$ p(x) = x^5+4x^4+9x^3+6x^2+9x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -0.0805+1.1216i\\[1 em]x_3 &= -0.0805-1.1216i\\[1 em]x_4 &= -1.9195+1.8529i\\[1 em]x_5 &= -1.9195-1.8529i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^5+4x^4+9x^3+6x^2+9x $ and solve two separate equations:

$$ \begin{aligned} x^5+4x^4+9x^3+6x^2+9x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^4+4x^3+9x^2+6x+9 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^4+4x^3+9x^2+6x+9 & = 0 \end{aligned} $$

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

Step 2:

Polynomial $ x^4+4x^3+9x^2+6x+9 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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