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!
 A: 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
$$2x^3-9x^2-10x+13=(2x+3)(x^2-6x+4)+1=(2x+3)(0)+1=0+1=1$$
A: Note $x-3=\sqrt{5}$. By successive synthetic division, the polynomial can be written as $$2(x-3)^3+9(x-3)^2-10(x-3)-44.$$ Plugging $\sqrt{5}$ for $x-3$ shows that the answer is $1$.
A: Build a polynomial that has $3+\sqrt5$ as a root, such as
$$x^2-6x+4$$ (using $3-\sqrt5$ for the other root).
Then by long division
$$(2x^3-9x^2-10x+13) \bmod (x^2-6x+4)=1.$$
A: We have
$$
(2x^3-9x^2-10x+13)-1=(x^2 - 6x + 4)(2x + 3)=0,
$$
since $x=3+\sqrt{5}$ satisfies the quadratic equation $x^2-6x+4=0$.
A: if $5 = (x-3)^2$ then $-10x = -2x(x-3)^2 = -2x^3+12x^2-18x$
Substituting this into $2x^3 - 9x^2 - 10x + 13$
we have
$$\begin{align}
2x^3 - 9x^2 +(-2x^3+12x^2-18x) + 13 &= 3x^2-18x+13\\
&= 3x^2-18x+27-14\\
&= 3(x^2-6x+9)-14\\
&= 3(x-3)^2-14\\
&= 3\cdot 5-14\\
&= 15-14\\
&= 1\\
\end{align}$$
A: I'm a month late to this, but just straight away plugging-in the expression can work, if you have the knowledge of the algebraic number you're dealing with.
What do I mean by this? Take;
$$x=3+\sqrt{5}=2\left(\frac{1+\sqrt{5}}{2}\right)^2=2\varphi^2$$
$\varphi$ is a well-known quantity in mathematics, called the 'golden ratio' And has the following properties;
$$\varphi^2=\varphi+1$$
$$1/\varphi=\varphi-1$$
$$\varphi^n=\varphi^{n-1}+\varphi^{n-2}$$
$$\varphi^n=F_{n}+F_{n+1}\varphi$$
Where $F_n$ is the $n$-th Fibonacci number ($F_0=0$). A very simple understanding of fibonacci sequences can be found here
So getting to the original question we get;
$$\begin{align}
2x^{3}-9x^{2}-10x+13 & = 16φ^{6}-36φ^{4}-20φ^{2}+13 \\
 & = 16\left(F_{6}+F_{7}φ\right)-36\left(F_{4}+F_{5}φ\right)-20\left(1+φ\right)+13 \\ 
 & = 16\left(5+8φ\right)-36\left(2+3φ\right)-20\left(1+φ\right)+13 \\
 & = \left(80-72-20+13\right)+φ\left(128-108-20\right) \\ 
 & = 1
\end{align}$$
This time, we were lucky that we got golden-ratio involved to simply expressions rather easily. This might not be the case in other questions, where you'll need to figure it out by other methods.
