Question
$$(2q^9+3q^7)(-6q^2+9) = 0$$
Answer
$$ q = 0 $$
Explanation
$$ \begin{aligned} (2q^9+3q^7)(-6q^2+9) &= 0&& \text{simplify left side} \\[1 em]-12q^{11}+18q^9-18q^9+27q^7 &= 0&& \\[1 em]-12q^{11}+18q^9-18q^9+27q^7 &= 0&& \\[1 em]-12q^{11}+27q^7 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -12x^{11}+27x^{7} = 0 } $, first we need to factor our $ x^7 $.
$$ -12x^{11}+27x^{7} = x^7 \left( -12x^{4}+27 \right) $$$ x = 0 $ is a root of multiplicity $ 7 $.
The remaining roots can be found by solving equation $ -12x^{4}+27 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
This page was created using
Polynomial Equations Solver