The remainder theorem says (in words):

If you have a polynomial function p(x) = (some terms in powers of x)
Then if you want to find p(c), then

Instead of substituting c for x in p(x), you can get the same answer
by performing synthetic division with the polynomial with c on the
far left of the synthetic divison, and taking ONLY the REMAINDER.


We put c = ½ to the left of the synthetic division   


  ½ | 8  6  2
    |________

Bring down the 8 below the bottom line:

  ½ | 8  6  2
    |________
      8

Multiply the 8 times the ½ getting (8)(½) = 4
and write 4 above the line and to the right of the 8,
underneath the 6:

  ½ | 8  6  2
    |    4    
      8

Add 6+4 getting 10, and write 10 below to line underneath
the 4 to the right of the 8:

  ½ | 8  6  2
    |    4   
      8 10

Multiply the 10 times the 0.5 getting (10)(½) = 5
and write 5 above the line and to the right of the 10,
underneath the 2, beside the 4:

  ½ | 8  6  2
    |    4  5
      8 10 

Add 2+5 getting 7, and write 7 below to line underneath
the 5 to the right of the 10:

  ½ | 8  6  2
    |    4  5
      8 10  7

So p(½) = the last number on the bottom is the remainder, 
which is 7. and that's equal to p(½)

Check by substituting ½ for x in 







So you see that you get the same answer 7 when you do synthetic
division with ½ on the far left as you get when you substitute ½
for x in the polynomial.  It's usually easier to use the remainder
theorem than to substitute -- especially when the polynomial has 
lots of terms.  This one only has 3 terms so the remainder theorem
is not that time-saving on this one, but it is when the polynomial
has more terms.

Edwin