Question
Find roots of polynomial $$ p(x) = 4x^4-x^3-7x^2 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.4538\\[1 em]x_3 &= -1.2038 \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x^2 }$ from $ 4x^4-x^3-7x^2 $ and solve two separate equations:
$$ \begin{aligned} 4x^4-x^3-7x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( 4x^2-x-7 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ 4x^2-x-7 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 4x^2-x-7 = 0 $ are: $ x = \dfrac{ 1 }{ 8 }-\dfrac{\sqrt{ 113 }}{ 8 } ~ \text{and} ~ x = \dfrac{ 1 }{ 8 }+\dfrac{\sqrt{ 113 }}{ 8 }$.
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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