Find roots of polynomial $$ p(x) = 3x^4+6x^3-7x^2 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.8257\\[1 em]x_3 &= -2.8257 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x^2 }$ from $ 3x^4+6x^3-7x^2 $ and solve two separate equations:

$$ \begin{aligned} 3x^4+6x^3-7x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( 3x^2+6x-7 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ 3x^2+6x-7 & = 0 \end{aligned} $$

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

Step 2:

The solutions of $ 3x^2+6x-7 = 0 $ are: $ x = -1-\dfrac{\sqrt{ 30 }}{ 3 } ~ \text{and} ~ x = -1+\dfrac{\sqrt{ 30 }}{ 3 }$.

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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