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Since x³+6x²+kx+12 is divisible by (x+4),
if we divide by x+4 by synthetic division
we should have a remainder of 0.
To divide by x+4 by synthetic division we
use the opposite signed version -4
-4|1 6 k 12
| -4 -8 -4k+32
1 2 k-8 -4k+44
The remainder must = 0
-4k+44 = 0
-4k = -44
k = 11
Edwin