Find roots of polynomial $$ p(x) = x+7x^2+9x^3+6x^4+5x^5 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -0.1804\\[1 em]x_3 &= -0.7766\\[1 em]x_4 &= -0.1215+1.1885i\\[1 em]x_5 &= -0.1215-1.1885i \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} x+7x^2+9x^3+6x^4+5x^5 & = 0\\[1 em] 5x^5+6x^4+9x^3+7x^2+x & = 0 \end{aligned} $$

Step 2:

Factor out $ \color{blue}{ x }$ from $ 5x^5+6x^4+9x^3+7x^2+x $ and solve two separate equations:

$$ \begin{aligned} 5x^5+6x^4+9x^3+7x^2+x & = 0\\[1 em] \color{blue}{ x }\cdot ( 5x^4+6x^3+9x^2+7x+1 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 5x^4+6x^3+9x^2+7x+1 & = 0 \end{aligned} $$

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

Step 3:

Polynomial $ 5x^4+6x^3+9x^2+7x+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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