jmc

Because $P(x)$ has three roots, if $Q_1(x)=x^2+(k-29)x-k$ and $Q_2(x)=2x^2+(2k-43)x+k$ are both factors of $P(x)$, then they must have a common root $r$. Then $Q_1(r)=Q_2(r)=0$, and $m{Q}_1(r)+n{Q}_2(r)=0$, for any two constants $m$ and $n$. Taking $m=2$ and $n=-1$ yields the equation $$2(r^2+(k-29)r-k)-(2r^2+(2k-43)r+k)=-3k-15r=0.$$Then $15r+3k=0$, so $r=\frac{-k}{5}$. Thus $$Q_1(r)=\frac{k^2}{25}-(k-29)\left(\frac{k}{5}\right)-k=0,$$which is equivalent to $4k^2-120k=0$, whose roots are $k=30$ and 0. When $k=30$, $Q_1(x)=x^2+x-30$ and $Q_2(x)=2x^2+17x+30$, and both polynomials are factors of $P(x)=(x+6)(x-5)(2x+5)$. Thus the requested value of $k$ is $\boxed{30}$.