We can use synthetic division to find out the remainders in terms of k.
The remainder for the first polynomial is 8k-1 and for the second polynomial k+6 (see later).
So these remainders are the same and 8k-1=k+6; 7k=7 so k=1.
2 | k 3 ......... 0..... -13
.....k 2k ....6+4k 12+8k
.....k 3+2k 6+4k | -1+8k Rem: 8k-1
2 | 2 0 -5 k
.....2 4 8 6
.....2 4 3 | k+6 Rem: k+6
The remainder is 7 in each case.