I already solved it for you here:

http://www.algebra.com/tutors/students/your-answer.mpl?question=978038

Maybe you thought that since I used long division instead of synthetic
division that that was not using the remainder theorem.  But synthetic
division is only a shorthand form of long division.  You can actually use
synthetic division.  You'll get the same thing.  I thought it might
confuse you since there are two variables.  OK, I'll do it with synthetic
division:

a⁴-7a²b²+k⁴

You have to put the b's in for they are coefficients of the a's.  Also you
must insert 0 terms for the missing powers of a

3b | 1  0b  -7b²  0b³  kb⁴
   |    3b   9b²  6b³ 18b⁴  
     1  3b   2b2  6b3 (kb⁴+18b⁴) 

The remainder must = 0 so that a-3b 
will be a factor of a⁴-7a²b²+kb⁴.

kb⁴+18b⁴ = 0
(k+18)b⁴ = 0
k+18=0; b⁴=0
 k=-18;  b=0  

There is a trivial case for b=0 and k is any number, 
which we ignore.

So k = -18

a⁴-7a²b²-18b⁴

(a²+2b²)(a²-9b²)

(a²+2b²)(a-3b)(a+3b)    <--- complete factorization

Edwin