Find roots of polynomial $$ p(x) = 6y^4+11y^2+4y $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 0\\[1 em]y_2 &= -0.3419\\[1 em]y_3 &= 0.1709+1.386i\\[1 em]y_4 &= 0.1709-1.386i \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ y }$ from $ 6y^4+11y^2+4y $ and solve two separate equations:
$$ \begin{aligned} 6y^4+11y^2+4y & = 0\\[1 em] \color{blue}{ y }\cdot ( 6y^3+11y+4 ) & = 0 \\[1 em] \color{blue}{ y = 0} ~~ \text{or} ~~ 6y^3+11y+4 & = 0 \end{aligned} $$One solution is $ \color{blue}{ y = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ 6y^3+11y+4 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.
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