x^2+kx+4 with k determined by a 6 sided die

Question

I need to know all the possibilities so I can know what the probability the answer will be factorable

Answer 1

I k can be any value between 1 and 6 so let’s see which values of k allow zeroes for the expression.

For the expression to have only real zeroes, k²-16≥0, so (k-4)(k+4)≥0. Therefore k≥4. k can’t be negative since a die produces only positive outcomes. The probability that k≥4 means that k can be 4, 5 or 6. This has a probability of 3/6=½. When k=4, factors are (x+2)(x+2).

When k=5, factors are (x+4)(x+1).

When k=6, factors are irrational, so perhaps we can't include them. Therefore there are only 2 values of k out of 6 that allow us to factorise rationally. The probability is 2/6=⅓.

If we allow irrational factors the probability is 3/6=½.