How to find values of k for which there will be 3 solutions

x^3 + kx - 4x - 2k = 0, therefore (x-2) is a factor and x=2 is a solution.

Find the values of k for which this equation will have 3 real solutions. (answer is k<1)
I think I need to do something with the descriminant, but I'm totally lost right now.

If someone could take me through the steps involved, it would be much appreciated. Thanks.

Factor out (x-2) and then you have a quadratic. Look at the discriminant of the quadratic term, and find those k for which it's > 0.

x^3 + kx - 4x - 2k
= (x^3 - 4x) + kx - 2k
= (x)(x^2 -4) + k(x-2)
= x(x-2)(x+2) + k(x-2)
=(x-2)[ (x)(x+2) + k ]
= (x-2)(x^2 +2x + k) = 0

Look at the discriminant of the quadratic term. Its 2^2 - 4(1)k = 4-4k
For 4-4k > 0
4 > 4k
1>k

So yes, answer is k<1.