Solution:

If $Q_1$ and $Q_2$ had no common root, their product — of degree $4$ — would divide the cubic $P(x),$ which is impossible. So they share a root $r,$ and $2Q_1(r) - Q_2(r) = 0.$ Computing, $2Q_1(x) - Q_2(x) = -15x - 3k,$ so $r = -\frac{k}{5}.$

Substituting into $Q_1(r) = 0$ gives $\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0;$ multiplying by $25$ and simplifying yields $-4k^2 + 120k = 0,$ so $k = 0$ or $k = 30.$

For $k = 30,$ $Q_1(x) = x^2 + x - 30 = (x + 6)(x - 5)$ and $Q_2(x) = 2x^2 + 17x + 30 = (x + 6)(2x + 5),$ and both divide $P(x) = (x + 6)(x - 5)(2x + 5).$ The largest value is $30.$