A Golden Rectangle Emerged!

If $x=\dfrac {3+\sqrt{5}}{2}$ , then it is the root of $x^2 - 3x +1 =0 \quad \Rightarrow x^2 = 3x -1$

Therefore,

$x^4 - x^3 - 6x^2 + 9x -4 = (3x-1)^2 - x(3x-1) - 6(3x-1) + 9x - 4$

$= 9x^2 - 6x + 1 - 3x^2 + x - 18x + 6 + 9x - 4 = 6x^2 - 14x + 3$

$= 6(3x-1) - 14x + 3 = 18x - 6 - 14x + 3 = 4x - 3$

$= 4 \left( \dfrac {3+\sqrt{5}}{2} \right) - 3 =6 + 2\sqrt{5} - 3 = \boxed {3+2\sqrt{5}}$

Since I couldn't manage to factorize the given expression, I solved it in a much lengthy way. Take the value that we are about to find as $S$ and we will focus solely on the value at $x=\dfrac{3+\sqrt{5}}{2}$ . We have,

$x=\dfrac{3+\sqrt{5}}{2} \\ \implies (2x-3)^2=5 \\ \implies 4x^2-12x+4=0 \\ \implies \boxed{x^2-3x+1=0} \\ \implies x^4-3x^3+x^2=0 \quad \text{[Multiplying both sides by } x^2 \text{ since } x\neq 0\text{ ]} \\ \implies \boxed{x^4=3x^3-x^2}$

Now, using these two results, we can reduce the degree of S and factorize it and further simplify as,

$S=x^4-x^3-6x^2+9x-4 \\ \implies S=3x^3-x^2-x^3-6x^2+9x-4 \\ \implies S=2x^3-7x^2+9x-4 \\ \implies S=(x-1)(2x^2-5x+4) \\ \implies S=(x-1)(2(x^2-3x+1)+(x+2)) \\ \implies S=(x-1)(x+2) \\ \implies S=x^2+x-2 \\ \implies S=(x^2-3x+1)+(4x-3) \\ \implies \boxed{S=4x-3}$

Now, $S$ has been simplified much and all you have to do is put the value of $x$ there and get the answer as $\boxed{3+2\sqrt{5}}$