Interval for Sum of Distinct Real Roots of Polynomial

To determine the interval in which the real number $s$ lies, we analyze the polynomial function:
$f(x) = 4x^3 + 3x^2 + 2x + 1$

Step 1: Determine the Number of Real Roots

We first find the derivative of $f(x)$:
$f'(x) = 12x^2 + 6x + 2 = 2(6x^2 + 3x + 1)$

The discriminant of the quadratic factor $6x^2 + 3x + 1$ is:
$\Delta = 3^2 - 4(6)(1) = 9 - 24 = -15 < 0$

Since the leading coefficient is positive ($6 > 0$) and the discriminant is negative ($\Delta < 0$), $f'(x) > 0$ for all $x \in \mathbb{R}$.

Thus, $f(x)$ is a strictly increasing function on $\mathbb{R}$, which implies that $f(x) = 0$ has exactly one real root, $s$. Consequently, the sum of all distinct real roots is simply $s$.

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Step 2: Locate the Root using the Intermediate Value Theorem

We evaluate $f(x)$ at $x = -\frac{1}{2}$ and $x = -\frac{3}{4}$:

1. For $x = -\frac{1}{2}$:
   $f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) + 1$
   $f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) - 1 + 1 = -\frac{1}{2} + \frac{3}{4} = \frac{1}{4} > 0$

2. For $x = -\frac{3}{4}$:
   $f\left(-\frac{3}{4}\right) = 4\left(-\frac{3}{4}\right)^3 + 3\left(-\frac{3}{4}\right)^2 + 2\left(-\frac{3}{4}\right) + 1$
   $f\left(-\frac{3}{4}\right) = 4\left(-\frac{27}{64}\right) + 3\left(\frac{9}{16}\right) - \frac{3}{2} + 1 = -\frac{27}{16} + \frac{27}{16} - \frac{3}{2} + 1 = -\frac{1}{2} < 0$

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Step 3: Conclusion

Since $f(x)$ is continuous on $\mathbb{R}$, $f\left(-\frac{3}{4}\right) < 0$, and $f\left(-\frac{1}{2}\right) > 0$, by the Intermediate Value Theorem, the unique real root $s$ must lie in the interval:
$s \in \left(-\frac{3}{4}, -\frac{1}{2}\right)$

Correct Answer: C