Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$? Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$?
The question is mentioned in my math olympiad. Please explain how to solve the problem. I have factorised the equation to $$-x^2+1, x-6x, -5x +1.$$
I am only in year 6.

If $c$ is a root of some polynomial, then $x - c$ is a a factor of that polynomial. Furthermore, if the polynomial has degree $n$, then it has at most $n$ roots. So $$x^3 - 6x^2 + 5x + 1 = (x - r)(x - s)(x - t)$$ Thus, 
\begin{align*}
(2 - r)(2 - s)(2 - t) &= 2^3 -6(2^2) + 5(2) + 1 \\
&= 8 - 6(4) + 10 + 1 \\
&= 8 - 24 + 10 + 1 \\
&= -5
\end{align*}
