Evaluating $2x^3-9x^2-10x+13$ for $x=3+\sqrt{5}$. Is there an efficient way to tell that this reduces to $1$? I was helping my sibling with a math problem from a past year paper for a competitive exam. It requires us to evaluate this cubic expression for a given value of $x$ which has an $a+b$ form where $b$ is a square root, as shown:

I essentially plugged in $x=a+b$ in the expression and expanded each term and finally got the answer as $1$. But this does not seem like the fastest way to do this, especially because there is only 1 minute given to solve each multiple choice question.
Is there a better way to reduce the original expression that gives the answer as $1$ or suggests that the square root term is going to evaluate to $0$?
Thanks!

Let $x=3+\sqrt{5}$.$\;$Then
\begin{align*}
&
x=3+\sqrt{5}\\[4pt]
\implies\;&
x-3=\sqrt{5}\\[4pt]
\implies\;&
(x-3)^2=5\\[4pt]
\implies\;&
x^2-6x+9=5\\[4pt]
\implies\;&
x^2-6x+4=0\\[4pt]
\end{align*}
Dividing $2t^3-9t^2-10t+13$ by $t^2-6t+4$ by polynomial long division, we get
$$2t^3-9t^2-10t+13=(2t+3)(t^2-6t+4)+1$$
hence
$$2x^3-9x^2-10x+13=(2x+3)(x^2-6x+4)+1=(2x+3)(0)+1=0+1=1$$
