Question
$$\frac{3x^3+28x^2-8x-9}{3}x^2 = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} \frac{3x^3+28x^2-8x-9}{3}x^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3 \cdot \frac{3x^3+28x^2-8x-9}{3}x^2 &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]3x^5+28x^4-8x^3-9x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{5}+28x^{4}-8x^{3}-9x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 3x^{5}+28x^{4}-8x^{3}-9x^{2} = x^2 \left( 3x^{3}+28x^{2}-8x-9 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 3x^{3}+28x^{2}-8x-9 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Polynomial Equations Solver