Problem (Algebra):
The four zeros of the polynomial $x^{4}+j x^{2}+k x+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$.

Solution:
Let the four zeros be $p \leq q \leq r \leq s$. The coefficient of $x^{3}$ is 0 , so $p+q+r+s=0$. The mean of four numbers in arithmetic progression is the mean of the middle two numbers, so $q=-r$. Then the common difference is $r-q=r-(-r)=2 r$, so $s=r+2 r=3 r$ and $p=q-2 r=-3 r$. Therefore the four zeros are $-3 r,-r, r, 3 r$. The product of



the zeros is $9 r^{4}$; referring to the original polynomial and using the product of roots formula gives $9 r^{4}=225$. Thus $r=\sqrt{5}$, the zeros are $-3 \sqrt{5},-\sqrt{5}, \sqrt{5}, 3 \sqrt{5}$, and the polynomial can be factored as $(x-\sqrt{5})(x+\sqrt{5})(x-3 \sqrt{5})(x+3 \sqrt{5})$. Expanding this product yields $\left(x^{2}-5\right)\left(x^{2}-45\right)=x^{4}-50 x^{2}+225$, so $j=-50$.

Proceed as in the original solution, finding the values $-3 \sqrt{5},-\sqrt{5}, \sqrt{5}$, and $3 \sqrt{5}$ for the zeros. By the sums and products of roots formulas, the coefficient of $x^{2}$ is the sum of all six possible products of pairs of roots:

$$
(-3 \sqrt{5})(-\sqrt{5})+(-3 \sqrt{5})(\sqrt{5})+(-3 \sqrt{5})(3 \sqrt{5})+(-\sqrt{5})(\sqrt{5})+(-\sqrt{5})(3 \sqrt{5})+(\sqrt{5})(3 \sqrt{5})
$$

Observing that some of these terms will cancel yields the simpler expression

$$
(-3 \sqrt{5})(3 \sqrt{5})+(-\sqrt{5})(\sqrt{5})=-45+-5=-50
$$

Final answer: -50