1. We say that a polynomial $p(x)$, with real coefficients, is Almeriense if it has the form

$$
p(x)=x^{3}+a x^{2}+b x+a
$$

and its three roots are positive real numbers in arithmetic progression. Find all Almeriense polynomials such that $p(7 / 4)=0$.
Solution. Let $\alpha \leq \beta \leq \gamma$ be the roots of the polynomial. From the condition of being in arithmetic progression, there exists a non-negative real number $\delta$ such that $\alpha=\beta-\delta$ and $\gamma=\beta+\delta$. On the other hand, using the Cardano-Viète formulas, we have

$$
\alpha \beta \gamma=-a=\alpha+\beta+\gamma
$$

which is equivalent to

$$
\beta\left(\beta^{2}-\delta^{2}\right)=3 \beta
$$

Since the roots are distinct from zero, we finally arrive at

$$
(\beta-\delta)(\beta+\delta)=\beta^{2}-\delta^{2}=3
$$

We need to analyze the three possible cases:

(a) If $\beta-\delta=7 / 4$, then $\beta+\delta=12 / 7$. From this, we have $\beta=97 / 56$ and $\delta=-1 / 56$, which contradicts the hypothesis that $\delta \geq 0$ (alternatively, it is not possible that $\beta+\delta<\beta-\delta$).

(b) If $\beta=7 / 4$, then $\delta^{2}=1 / 4$. Therefore, the roots of the polynomial are $(3 / 2,7 / 4,2)$, and we obtain

$$
p(x)=(x-3 / 2)(x-7 / 4)(x-2)=x^{3}-\frac{21}{4} x^{2}+\frac{73}{8} x-\frac{21}{4}
$$

(c) If $\beta+\delta=7 / 4$, then $\beta-\delta=12 / 7$. From this, we have $\beta=97 / 56$ and $\delta=1 / 56$. Therefore, the roots of the polynomial are $(12 / 7,97 / 56,7 / 4)$, and we obtain

$$
p(x)=(x-12 / 7)(x-97 / 56)(x-7 / 4)=x^{3}-\frac{291}{56} x^{2}+\frac{14113}{1568} x-\frac{291}{56}
$$