, 2014

Solution:
Answer: $\pm \frac{9}{4}$ (OR $\frac{9}{4},-\frac{9}{4}$ OR $\left.-\frac{9}{4}, \frac{9}{4}\right)$ OR $\pm 2 \frac{1}{4}$ OR $\pm 2.25$

Any real quartic has an even number of real roots with multiplicity, so there exists real $r$ such that $x^{4}+k x^{3}+x^{2}+4 k x+16$ either takes the form $(x+r)^{4}$ (clearly impossible) or $(x+r)^{2}\left(x^{2}+a x+b\right)$ for some real $a, b$ with $a^{2}<4 b$. Clearly $r \neq 0$, so $b=\frac{16}{r^{2}}$ and $4 k=4(k)$ yields $\frac{32}{r}+a r^{2}=4(2 r+a) \Longrightarrow a\left(r^{2}-4\right)=8 \frac{r^{2}-4}{r}$. Yet $a \neq \frac{8}{r}$ (or else $a^{2}=4 b$ ), so $r^{2}=4$, and $1=r^{2}+2 r a+\frac{16}{r^{2}} \Longrightarrow a=\frac{-7}{2 r}$. Thus $k=2 r-\frac{7}{2 r}= \pm \frac{9}{4}$ (since $r= \pm 2$ ).

It is easy to check that $k=\frac{9}{4}$ works, since $x^{4}+(9 / 4) x^{3}+x^{2}+4(9 / 4) x+16=\frac{1}{4}(x+2)^{2}\left(4 x^{2}-7 x+16\right)$. The polynomial given by $k=-\frac{9}{4}$ is just $\frac{1}{4}(-x+2)^{2}\left(4 x^{2}+7 x+16\right)$.

Alternate solution: $x^{4}+k x^{3}+x^{2}+4 k x+16=\left(x^{2}+\frac{k}{2} x+4\right)^{2}+\left(1-8-\frac{k^{2}}{4}\right) x^{2}$, so for some $\epsilon \in\{-1,1\}$, $2 x^{2}+\left(k-\epsilon \sqrt{k^{2}+28}\right) x+8$ has a single real root and thus takes the form $2(x+r)^{2}$ (using the same notation as above). But then $\left(k-\epsilon \sqrt{k^{2}+28}\right)^{2}=4(2)(8)=8^{2}$, so we conclude that $(k \pm 8)^{2}=\left(\epsilon \sqrt{k^{2}+28}\right)^{2}$ and $k= \pm\left(4-\frac{7}{4}\right)= \pm \frac{9}{4}$.