Solution:
$f(-1) + f(5) = 202$
Step by step solution:
Let $g(x) = f(x) + ax^2 + bx + c$
Find $a,\, b,\, c$ so that $x = 1;\, x = 2,\, x = 3$ are solutions of $g(x)$
We get:
$g(1) = g(2) = g(3) = 0$
$\Leftrightarrow \begin{cases}f(1) + a + b + c = 0\\f(2) + 4a + 2b + c = 0\\f(3) + 9a + 3b + c = 0\end{cases}$
$\Leftrightarrow \begin{cases} a + b + c = -5\\4a + 2b + c = -11\\9a + 3b + c = -21\end{cases}$
$\Leftrightarrow \begin{cases}a = -2\\b = 0\\c = -3\end{cases}$
Therefore: $g(x) = f(x) - 2x^2 - 3$
$\Leftrightarrow f(x) = g(x) + 2x^2 + 3$
$\Leftrightarrow f(x) = (x-1)(x-2)(x-3)(x-x_o) + 2x^2 + 3$
$\Rightarrow \begin{cases}f(-1) = 24(1+x_o) + 5\\f(5) = 24(5 - x_o) + 53\end{cases}$
So:
$f(-1) + f(5) = 24(1+x_o) + 5 + 24(5 - x_o) + 53$
$\Rightarrow f(-1) + f(5) = 29 + 24x_o + 173 - 24x_o$
$\Rightarrow f(-1) + f(5) = 202$
