Question 978568
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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:

a&#8308;-7a²b²+k&#8308;

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  kb&#8308;
   |<u>    3b   9b²  6b³ 18b&#8308;</u>  
     1  3b   2b2  6b3 (kb&#8308;+18b&#8308;) 

The remainder must = 0 so that a-3b 
will be a factor of a&#8308;-7a²b²+kb&#8308;.

kb&#8308;+18b&#8308; = 0
(k+18)b&#8308; = 0
k+18=0; b&#8308;=0
 k=-18;  b=0  

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

So k = -18

a&#8308;-7a²b²-18b&#8308;

(a²+2b²)(a²-9b²)

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

Edwin</pre>