Question 1074901
<pre><b>
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
  |<u>     -4    -8  -4k+32</u> 
   1     2   k-8  -4k+44

The remainder must = 0

                  -4k+44 = 0
                     -4k = -44
                       k = 11

Edwin</pre></b>