The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.3177\\[1 em]x_3 &= -1.3177\\[1 em]x_4 &= -1.8066\\[1 em]x_5 &= 1.8066 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x }$ from $ 3x^5-15x^3+17x $ and solve two separate equations:
$$ \begin{aligned} 3x^5-15x^3+17x & = 0\\[1 em] \color{blue}{ x }\cdot ( 3x^4-15x^2+17 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 3x^4-15x^2+17 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ 3x^4-15x^2+17 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.