I and Phi?

I made this problem to prove that this factoring trick can be used for polys of all degrees.

Let's look at the constant term. Using the identity, $\phi^n=\phi^{n-1}+\phi^{n-2}$ .

$210+315\phi=105+105\phi+105\phi+105+105\phi$

$\Longrightarrow 105\phi^2+105\phi^2+105\phi=105\phi^2+105\phi^3=105\phi^4$

Now we have

$105\phi^4+76i\phi^3x+34\phi^2x^2+4i\phi x^3+x^4=0$

Using this . We extract $\phi$ and $i$ inversely with $x$ (aka, as the power of $x$ decreases, we factor out larger and larger powers of $i$ and $\phi$ ) and are left with

$105-76x-34x^2+4x^3+x^4=0$

Remainder theorem tells us that $x=1$ is a root.

$(x-1)(x^3+5x^2-29x-105)=0$

Once again, remainder theorem/rational roots tells us that $x=5$ is a root.

$(x-1)(x-5)(x^2+10x+21)=0$

$(x-1)(x-5)(x+3)(x+7)=0$

Factoring back in the $i$ and $\phi$

$(x-i\phi)(x-5i\phi)(x+3i\phi)(x+7i\phi)=0$

Thus $a_1=i\phi,~a_2=-3i\phi,~a_3=5i\phi,~a_4=-7i\phi$ and

$\dfrac{i\phi-3i\phi-5i\phi}{-7i\phi}=\boxed{1}$