find the value of k if 2x^4+3x^3+2kx^2+3x+6 is exactly divisible by (x+2)
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put x=-2 and equate term to zero
2×(-2)^4+3(-2)^3+2k(-2)^2+3(-2)+6=0
(2×16)+(3×-8)+(2k×4)-6+6=0
32-24+8k-6+6=0
8+8k=0
8k=-8
k =-1
2×(-2)^4+3(-2)^3+2k(-2)^2+3(-2)+6=0
(2×16)+(3×-8)+(2k×4)-6+6=0
32-24+8k-6+6=0
8+8k=0
8k=-8
k =-1
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