Step 1: Analyze the function $f(x)$ and its roots.

The given polynomial is $f(x) = 1 + 2x + 3x^2 + 4x^3$. To find information about the roots of $f(x)$, we first examine its derivative, $f'(x)$. $f'(x) = \frac{d}{dx}(1 + 2x + 3x^2 + 4x^3) = 2 + 6x + 12x^2.$ To determine the nature of the roots of the quadratic equation $f'(x) = 0$, we calculate its discriminant, $\Delta$. $12x^2 + 6x + 2 = 0$ $\Delta = b^2 - 4ac = 6^2 - 4(12)(2) = 36 - 96 = -60.$ Since the discriminant is negative ($\Delta < 0$) and the leading coefficient (12) is positive, the quadratic $f'(x)$ is always positive for all real values of $x$. This means $f'(x) > 0$ for all $x \in \mathbb{R}$. Therefore, the function $f(x)$ is strictly increasing for all real $x$. A strictly increasing function can cross the x-axis at most once, which implies that $f(x)$ has exactly one real root. Let's call this root $\alpha$.

Step 2: Determine the value of $s$ and $t$.

According to the problem statement, $s$ is the sum of all distinct real roots of $f(x)$. Since there is only one real root, $\alpha$, we have $s = \alpha$. Then, $t = |s| = |\alpha|$. To find the interval in which $\alpha$ lies, we can test some values for $f(x)$: $f(0) = 1 + 0 + 0 + 0 = 1 > 0$. $f(-1) = 1 - 2 + 3 - 4 = -2 < 0$. Since $f(x)$ is continuous and $f(-1) < 0$ and $f(0) > 0$, by the Intermediate Value Theorem, the root $\alpha$ must lie between -1 and 0. So, $-1 < \alpha < 0$.

Step 3: Analyze the function $f'(x)$ for its monotonicity.

The question asks about the increasing/decreasing nature of the function $f'(x) = 12x^2 + 6x + 2$. To study the monotonicity of $f'(x)$, we need to find its derivative, which is $f''(x)$. $f''(x) = \frac{d}{dx}(12x^2 + 6x + 2) = 24x + 6.$ We find the critical point of $f'(x)$ by setting $f''(x) = 0$. $24x + 6 = 0 \implies 24x = -6 \implies x = -\frac{6}{24} = -\frac{1}{4}.$The function $f'(x)$ is a parabola opening upwards, and its minimum occurs at its vertex, $x = -1/4$. The sign of $f''(x)$ determines the monotonicity of $f'(x)$:

• If $x < -1/4$, $f''(x) = 24x + 6 < 0$, which means $f'(x)$ is decreasing.
• If $x > -1/4$, $f''(x) = 24x + 6 > 0$, which means $f'(x)$ is increasing.

Step 4: Relate the monotonicity of $f'(x)$ to the interval $(-t, t)$.

The interval of interest is $(-t, t)$, where $t = |\alpha|$. We need to determine if the critical point $x = -1/4$ lies inside this interval. This requires comparing $t$ with $1/4$. Let's evaluate $f(x)$ at $x = -1/4$: $f\left(-\frac{1}{4}\right) = 1 + 2\left(-\frac{1}{4}\right) + 3\left(-\frac{1}{4}\right)^2 + 4\left(-\frac{1}{4}\right)^3$ $= 1 - \frac{1}{2} + 3\left(\frac{1}{16}\right) + 4\left(-\frac{1}{64}\right)$ $= \frac{1}{2} + \frac{3}{16} - \frac{4}{64} = \frac{1}{2} + \frac{3}{16} - \frac{1}{16} = \frac{1}{2} + \frac{2}{16} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}.$Since $f(x)$ is a strictly increasing function, and we know $f(\alpha) = 0$ and $f(-1/4) = 5/8 > 0$, it must be that $\alpha < -1/4$. Taking the absolute value, we get $|\alpha| > |-1/4|$, which means $t > 1/4$. Since $t > 1/4$, the interval $(-t, t)$ contains the point $-1/4$. Specifically, $-t < -1/4 < t$. Now we can describe the behavior of $f'(x)$ on the interval $(-t, t)$. We split the interval at the critical point $x=-1/4$:

• In the interval $\left(-t, -\frac{1}{4}\right)$, we have $x < -1/4$, so $f'(x)$ is decreasing.
• In the interval $\left(-\frac{1}{4}, t\right)$, we have $x > -1/4$, so $f'(x)$ is increasing.

Step 5: Conclusion.

The function $f'(x)$ is decreasing in $\left( -t, - {1 \over 4} \right)$ and increasing in $\left( - {1 \over 4},t \right)$. This matches option B.

Final check of options: A: increasing in $\left( { - t, - {1 \over 4}} \right)$ and decreasing in $\left( { - {1 \over 4},t} \right)$ - Incorrect. B: decreasing in $\left( { - t, - {1 \over 4}} \right)$ and increasing in $\left( { - {1 \over 4},t} \right)$ - Correct. C: increasing in $(-t, t)$ - Incorrect. D: decreasing in $(-t, t)$ - Incorrect.