The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{\sqrt{ 65 }}{ 13 }\\[1 em]x_3 &= - \frac{\sqrt{ 65 }}{ 13 } \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 3x }$ from $ 39x^3-15x $ and solve two separate equations:
$$ \begin{aligned} 39x^3-15x & = 0\\[1 em] \color{blue}{ 3x }\cdot ( 13x^2-5 ) & = 0 \\[1 em] \color{blue}{ 3x = 0} ~~ \text{or} ~~ 13x^2-5 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 13x^2-5 = 0 $ are: $ x = - \dfrac{\sqrt{ 65 }}{ 13 } ~ \text{and} ~ x = \dfrac{\sqrt{ 65 }}{ 13 }$.
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