f(x)=x4+x3=16-8x-6x2 means f(x)=x4+x3 and f(x)=16-8x-6x2, therefore x4+x3=16-8x-6x2.
So x4+x3-16+8x+6x2=0,
x4+x3+6x2+8x-16=0.
Note that if x=1: 1+1+6+8-16=0, therefore x=1 is a zero of the polynomial.
Use synthetic division to divide by this zero:
1 | 1 1 6 8 -16
1 1 2 8 | 16
1 2 8 16 | 0 = x3+2x2+8x+16.
There may be another rational zero. If it exists, it will be a factor of 16 and it will be negative because this cubic can only factorise with positive factors (all signs are plus) of the type x+a, where a is positive, which means that the zeroes must be such that x+a=0, that is, x=-a. So let's start by plugging in x=-2:
-8+8-16+16=0, making x=-2 another zero. So we divide by this zero:
-2 | 1 2 8 16
1 -2 0 | -16
1 0 8 | 0 = x2+8 which has no real zeroes (only complex factors ±2i√2).
Assuming only real zeroes apply in this question, f(x)=x4+x3. Plug in the two zeroes f(x)=1+1=2, or f(x)=16-8=8.
The same zeroes can be plugged into f(x)=16-8x-6x2=16-8-6=2, or f(x)=16+16-24=8.