SOLUTION: Let p(x)=9x^3-36x^2-9x+36 a. Use synthetic division to find P(3) b. Divide p(x) by x-4 and write p(x)=(x-4)q(x)+r

We use the remainder theorem.  That says the remainder in any synthetic
division problem is exactly the same number as you would get if you 
substituted the number on the far left of the synthetic division into the
polynomial.  

3 | 9  -36   -9   36
  |     27  -27 -108 
    9   -9  -36  -72 

so -72 is the same number as you would get if you substituted the number on the
far left of the synthetic division, which is 3, into the polynomial.  And that
is what we call P(3).  So P(3) = -72
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b. Divide p(x) by x-4 and write p(x)=(x-4)q(x)+r

4 | 9  -36   -9   36
  |     36    0  -36 
    9    0   -9    0 

The numbers on the bottom row, except the last one determines the quotient
polynomial q(x) which is one degree lower than the original.  So the numbers on
the bottom of the synthetic division except the last one, 9  0 -9, determines
the quotient polynomial which is q(x)=9x³+0x²-9 or q(x)=9x²-9. r=0 and
so 
p(x)=(x-4)q(x)+r  becomes

p(x)=(x-4)(9x²-9)+0

Edwin