Find roots of polynomial $$ p(x) = x^5+x^3-6x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.4142\\[1 em]x_3 &= -1.4142\\[1 em]x_4 &= 1.7321i\\[1 em]x_5 &= -1.7321i \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ x^5+x^3-6x $ and solve two separate equations:
$$ \begin{aligned} x^5+x^3-6x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^4+x^2-6 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^4+x^2-6 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ x^4+x^2-6 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.
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