It is given that `x^3-7x+6` is divisible by `x^2-3x+k` .
Therefore `x^2-3x+k` is a factor of` x^3-7x+6` . If the other factor is (x+p) we can say that;
`x^3-7x+6 = (x^2-3x+k)(x+p)`
`x^3-7x+6 = x^3+x^2(p-3)+x(k-3p)+kp`
By considering each coefficient;
`x^3rarr 1 = 1`
`x^2rarr 0 = p-3` hence `p = 3`
`x rarr -7 = k-3p` hence `k = -7+3p = -7+3xx3 = 2`
So the value of k = 2. The correct answer is at B.