x^3-4kz^2-k^2x+4=0

Question

Find the integer value(s) ok k for which the given equation has rational roots and find the roots. (hint: use synthetic division)

Answer 1

I'm assuming z=x. 

x³-4kx²-k²x+4=x(x²-k²)-4(kx²-1).

From this it can be seen that putting k=1 gives us:

x(x²-1)-4(x²-1)=(x-4)(x²-1)=(x-4)(x-1)(x+1), so k=1 is a solution, yielding three roots: 4, 1, -1.

If we use one of these we can reduce the cubic to a quadratic through synthetic division:

1 | 1   -4k      -k²         4

     1     1     1-4k   | 1-4k-k²

     1 1-4k 1-4k-k² | 5-4k-k²=0=(5+k)(1-k), so k=1 (already identified) and k=-5.

The quadratic becomes x²-3k-4=0=(x-4)(x+1), when k=1, which is the solution from above.

It also becomes x²+21x-4=0, but this yields irrational roots. Now try rational zero=2

2 | 1   -4k      -k²             4

     1      2   4-8k | -2k²-16k

     1 2-4k -k²-8k | -2k²-16k+4=-2(k²+8k-2)=0, which has irrational roots.

So it appears that k=1 is the only value of k which produces rational roots.