Find roots of polynomial $$ p(x) = x^5-3x^4+\frac{1}{3}x^3-4x^2-5x+9 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.9074\\[1 em]x_2 &= -1.1988\\[1 em]x_3 &= 3.3241\\[1 em]x_4 &= -0.0164+1.5776i\\[1 em]x_5 &= -0.0164-1.5776i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 3 } $.
$$ \begin{aligned} x^5-3x^4+\frac{1}{3}x^3-4x^2-5x+9 & = 0 ~~~ / \cdot \color{blue}{ 3 } \\[1 em] 3x^5-9x^4+x^3-12x^2-15x+27 & = 0 \end{aligned} $$Step 2:
Polynomial $ 3x^5-9x^4+x^3-12x^2-15x+27 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.
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