Two routes to the same answer

First Solution

Let $x=3$ and replace in the first given equation, to obtain $\left( 3+3 \right) \left( 3+1 \right) f\left( 3 \right) =\left( 3+2 \right) \left( 3+4 \right) f\left( 3-1 \right)$ . Equalizing with the second given equation and solving, we get $f\left( 3 \right) =\frac { 35 }{ 24 } f\left( 2 \right) =f\left( 2 \right) +22$ , then

$\frac { 11 }{ 24 } f\left( 2 \right) =22\\ f\left( 2 \right) =48,\quad f\left( 3 \right) =70$ .

Now, let $x=4$ and replace in the first given equation, thus

$7\cdot 5\cdot f\left( 4 \right) =6\cdot 8\cdot f\left( 3 \right) \\ f\left( 4 \right) =96$ .

Second Solution

Note that $x=-2$ and $x=-4$ are roots of the polynomial $f\left( x \right)$ , hence, we can write this polynomial as $f\left( x \right) =a\cdot \left( x+2 \right) \left( x+4 \right)$ , where $a$ is a constant. From the second given equation we obtain $f\left( 3 \right) =35a=24a+22$ , therefore, $a=2$ , and, thus, $f\left( x \right) =2\left( x+2 \right) \left( x+4 \right)$ . Then, $f\left( 4 \right) =2\cdot 6\cdot 8=96$ .

The proof that this polynomial satisfies the given conditions is simple:

Knowing that $f\left( x \right) =2\left( x+2 \right) \left( x+4 \right)$ , then $f\left( x-1 \right) =2\left( x+1 \right) \left( x+3 \right)$ . Multiply these equations to obtain the first given equation.

P.S.: This is not the only polynomial that works, but it is the simplest to find.