Problem 1444

1. Let the roots of the polynomial $p(x)$ be $\{c, c+d, c+2d\}$ where $c$ and $d$ are positive real numbers. Since the roots are in arithmetic progression, we can use Vieta's formulas to express the coefficients of the polynomial in terms of $c$ and $d$.

2. By Vieta's formulas, the sum of the roots is equal to the coefficient of $x^2$ with a negative sign:
$$c + (c + d) + (c + 2d) = 3c + 3d = -a$$
The product of the roots taken one at a time is equal to the constant term:
$$c(c + d)(c + 2d) = -a$$

3. Given that $p\left(\frac{7}{4}\right) = 0$, one of the roots must be $\frac{7}{4}$. We consider two cases: $c = \frac{7}{4}$ and $c + d = \frac{7}{4}$.

4. **Case 1: $c = \frac{7}{4}$**
$$3c + 3d = -a \implies 3\left(\frac{7}{4}\right) + 3d = -a \implies \frac{21}{4} + 3d = -a$$
$$c(c + d)(c + 2d) = -a \implies \left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right)\left(\frac{7}{4} + 2d\right) = -a$$

5. **Case 2: $c + d = \frac{7}{4}$**
$$c + (c + d) + (c + 2d) = 3c + 3d = -a \implies 3c + 3\left(\frac{7}{4} - c\right) = -a \implies \frac{21}{4} = -a$$
$$c(c + d)(c + 2d) = -a \implies c\left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right) = -a$$

6. Solving for $d$ in both cases:
- For $c = \frac{7}{4}$:
$$3\left(\frac{7}{4}\right) + 3d = -a \implies \frac{21}{4} + 3d = -a$$
$$\left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right)\left(\frac{7}{4} + 2d\right) = -a$$
Solving these equations, we find $d = \frac{1}{6}$ and $c = \frac{3}{2}$.

- For $c + d = \frac{7}{4}$:
$$c + d = \frac{7}{4} \implies c = \frac{7}{4} - d$$
$$c\left(\frac{7}{4}\right)\left(\frac{7}{4} + d\right) = -a$$
Solving these equations, we find $d = \frac{1}{96}$ and $c = \frac{12}{7}$.

7. Constructing the polynomials for each case:
- For $c = \frac{3}{2}$ and $d = \frac{1}{6}$:
$$p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4}$$

- For $c = \frac{12}{7}$ and $d = \frac{1}{96}$:
$$p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}$$

The final answer is $\boxed{ p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4} }$ or $p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}$.