Find roots of polynomial $$ p(x) = x^4-x^3-5x^2 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 2.7913\\[1 em]x_3 &= -1.7913 \end{aligned} $$

Step 1:

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

$$ \begin{aligned} x^4-x^3-5x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( x^2-x-5 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ x^2-x-5 & = 0 \end{aligned} $$

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

Step 2:

The solutions of $ x^2-x-5 = 0 $ are: $ x = \dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 21 }}{ 2 } ~ \text{and} ~ x = \dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 21 }}{ 2 }$.

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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