Let me give two quick examples. Let $\mathbf{u}_1$ and $\mathbf{u}_2$ be a multivectors of $Cl_{3,0}(\mathbb{R})$:
$$ \begin{align} \mathbf{u}_1&=x e_1+y e_2\\ \mathbf{u}_2&=a+be_1 \end{align} $$
For $\mathbf{u}_1$, I can construct a polynomial (of degree 2) that erases all basis elements, simply by taking the geometric product:
$$ \mathbf{u}_1^2 = x^2+y^2 $$
For $\mathbf{u}_2$ this is also possible as follows:
$$ \begin{align} \mathbf{u}_2^2&=a^2+b^2+2abe_1\\ \mathbf{u}_2^2&=a(2a+2be_1)-a^2+b^2\\ \mathbf{u}_2^2&=a(2\mathbf{u})-a^2+b^2\\ \mathbf{u}_2^2-2a\mathbf{u}+a^2-b^2&=0 \end{align} $$
Of course, as a restriction, the original value of $\mathbf{u}_1$ and $\mathbf{u}_2$ must be a root of the polynomial. Bonus points if the produced polynomial is the lowest degree possible.
Is it possible to produce such a polynomial when we are talking about the most general multivector of say $Cl_{3,0}(\mathbb{R})$ or even $Cl_{3,1}(\mathbb{R})$:
$$ \mathbf{u}:=a\\ +t\gamma_0+x\gamma_1+y\gamma_2+z\gamma_3\\ +E_x\gamma_0\gamma_1+E_y\gamma_0\gamma_2+E_z\gamma_0\gamma_3\\ +B_x\gamma_2\gamma_3+B_y\gamma_1\gamma_3+B_z\gamma_1\gamma_2\\ +V_t\gamma_0+V_x\gamma_1+V_y\gamma_2+V_z\gamma_3\\ +b\gamma_0\gamma_1\gamma_2\gamma_3 $$
EDIT
Finding the roots of a complex number in $Cl_{3}(\mathbb{R})$ by solving a polynomial:
$$ \begin{align} \mathbf{u}:&=a+bI=a+b\sigma_x\sigma_y\sigma_z\\ \implies \mathbf{u}^2&=a^2-b^2+2abI\\ &=a(2a+2bI-a)-b^2\\ &=a(2\mathbf{u}-a)-b^2\\ &=2a\mathbf{u}-a^2-b^2\\ \implies \mathbf{u}^2-2a\mathbf{u}+a^2+b^2&=0\\ \implies \mathbf{u}&=a\pm\sqrt{-b^2}=a\pm ib \end{align} $$
Finding the eigenvalues of the matrix representation of $\mathbf{u}$:
$$ \begin{align} \mathbf{u}=a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&i}=\pmatrix{a+ib&0\\0&a+ib} \end{align} $$
The eigenvalues are the same: $a+ib$
The roots of the polynomial are not the same as the eigenvalues. Can I use the matrix representation to solve for the roots of the polynomial? This would be a different equation to solve than the equation of the characteristic polynomial?