If $a$ is a root of $x^2(x-5)+2$, then what is $[a^4]$? If $a$ is the largest root of $f(x)=x^2(x-5)+2$, then what is $[a^4]$ where $[\ ]$ is a Gauss bracket ? (i.e. $[x]$ is a largest integer not strictly greater than $x$)
Here $f(5)=2,\ f(4)=-14$. By considering $f'(x)$ we can know that $a$ is in an open interval $(4,5)$.
If $b_0=4$, then we define $b_n$ : $$ \frac{f(5)-  f(b_n) }{5-b_n } (b_{n+1} - 5) +f(5)  =0 \ (i.e. b_{n+1} = 5- \frac{2}{b_n^2} ) $$
Note that an increasing sequence $b_n$ s.t. $|\dfrac{b_{n}-b_{n+1} }{b_{n-1} - b_{n} } |\leq \frac{1}{12}$ goes to $a$ since $f''(x)>0$ on $[4,5]$. And $$ a\in [b_3,b_3+C],\ C = \frac{7}{8\cdot 11 \cdot (12)^2} $$
Here $[(b_3)^4] =[(b_3+C)^4]=584$ But this is a numerical way. How can we calculate $[a^4]$ in the highschool level ?

Combining two ideas from the comments.
Let $a,b,c$ be the zeros with $a$ the largest. By expanding we see that
$$f(x)f(-x)=4-20x^2+25x^4-x^6=-g(x^2),$$
where
$$g(x)=x^3-25x^2+20x-4$$
has $a^2,b^2$ and $c^2$ as its zeros. By the Vieta relations it follows that
$$
s_1:=a^2+b^2+c^2=25
$$
and the second symmetric polynomial
$$
s_2:=a^2b^2+a^2c^2+b^2c^2=20.
$$
Therefore
$$
a^4+b^4+c^4=s_1^2-2s_2=585.
$$
But $f(-3/4)<0<f(0)$ and $f(3/4)<0$, so $|b|$ and $|c|$ are both $<3/4$.
Therefore $0<b^4+c^4<1$ and hence
$$
584<a^4<585.
$$
