Question
Find roots of polynomial $$ p(x) = 118709x+100314x^2+999983x^3 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -\dfrac{ 50157 }{ 999983 }+0.3409i\\[1 em]x_3 &= -\dfrac{ 50157 }{ 999983 }-0.3409i \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} 118709x+100314x^2+999983x^3 & = 0\\[1 em] 999983x^3+100314x^2+118709x & = 0 \end{aligned} $$Step 2:
Factor out $ \color{blue}{ x }$ from $ 999983x^3+100314x^2+118709x $ and solve two separate equations:
$$ \begin{aligned} 999983x^3+100314x^2+118709x & = 0\\[1 em] \color{blue}{ x }\cdot ( 999983x^2+100314x+118709 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 999983x^2+100314x+118709 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 3:
The solutions of $ 999983x^2+100314x+118709 = 0 $ are: $ x = -\dfrac{ 50157 }{ 999983 }+0.3409i ~ \text{and} ~ x = -\dfrac{ 50157 }{ 999983 }-0.3409i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
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