Answer: \(k=3\), and the remaining quadratic factor has no real roots.

Answer: \(k=3\), and the line intersects the curve at only one point.

At an intersection,

\(kx+6=x^3-4x^2+3kx+2.\)

Substitute \(x=2\):

\(2k+6=8-16+6k+2.\)

So

\(2k+6=6k-6,\)

which gives

\(k=3.\)

Now substitute \(k=3\) into the intersection equation:

\(3x+6=x^3-4x^2+9x+2.\)

Hence

\(x^3-4x^2+6x-4=0.\)

Since \(x=2\) is a root, factorise:

\(x^3-4x^2+6x-4=(x-2)(x^2-2x+2).\)

The quadratic factor has discriminant

\((-2)^2-4(1)(2)=4-8=-4\lt 0.\)

So it has no real roots. Therefore the only real intersection occurs at \(x=2\), and the line intersects the curve at only one point.