The [polynomial](https://artofproblemsolving.com/wiki/index.php/Polynomial) $P(x)$ is [ cubic](https://artofproblemsolving.com/wiki/index.php/Cubic_polynomial).  What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both [factors](https://artofproblemsolving.com/wiki/index.php/Factor) of $P(x)$?
Solution 1
We can see that $Q_1$ and $Q_2$ must have a [root](https://artofproblemsolving.com/wiki/index.php/Root) in common for them to both be [factors](https://artofproblemsolving.com/wiki/index.php/Factor) of the same cubic.
Let this root be $a$.
We then know that $a$ is a root of
$Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$
, so $x = \frac{-k}{5}$.
We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get:
$\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$
or $k^{2}=30k$, so $k=30$ is the highest.
We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.

Solution 2
Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$.
Therefore, we can write four equations (and we have four [variables](https://artofproblemsolving.com/wiki/index.php/Variable)), $a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $am = -k$, and $an = \frac{k}{2}$. 
The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$. The last two equations show that $\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = \boxed{030}$.

Solution 3
Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations: 
\begin{align}  a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck  \end{align}
Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed {030}$.  
~ anellipticcurveoverq

Solution 4
Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$
Thus, we see that we have 
\[r^2 + (k - 29)r - k = 0,\]
\[2r^2 + (2k - 43)r + k = 0.\] Adding the two equations gives us \[3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.\] This can easily be checked to see that it does indeed work, and we're done! 
~Ilikeapos