Find roots of polynomial $$ p(x) = 48x^5+40x^4-36x^3-18x^2+4x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.1753\\[1 em]x_3 &= -0.5784\\[1 em]x_4 &= 0.7167\\[1 em]x_5 &= -1.147 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ 2x }$ from $ 48x^5+40x^4-36x^3-18x^2+4x $ and solve two separate equations:

$$ \begin{aligned} 48x^5+40x^4-36x^3-18x^2+4x & = 0\\[1 em] \color{blue}{ 2x }\cdot ( 24x^4+20x^3-18x^2-9x+2 ) & = 0 \\[1 em] \color{blue}{ 2x = 0} ~~ \text{or} ~~ 24x^4+20x^3-18x^2-9x+2 & = 0 \end{aligned} $$

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

Step 2:

Polynomial $ 24x^4+20x^3-18x^2-9x+2 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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