Number of Distinct Real Roots of Fourth Degree Polynomial

To find the number of distinct real roots of the fourth-degree polynomial equation:
$f(x) = x^4 - 4x^3 + 12x^2 + x - 1 = 0$

we can analyze the behavior and extrema of the function $f(x)$ using its derivatives.

Step 1: Compute the first and second derivatives

The first derivative of $f(x)$ is:
$f'(x) = \frac{d}{dx}\left(x^4 - 4x^3 + 12x^2 + x - 1\right) = 4x^3 - 12x^2 + 24x + 1$

The second derivative of $f(x)$ is:
$f''(x) = \frac{d}{dx}\left(4x^3 - 12x^2 + 24x + 1\right) = 12x^2 - 24x + 24$

Factoring out $12$, we get:
$f''(x) = 12(x^2 - 2x + 2) = 12\left((x - 1)^2 + 1\right)$

Step 2: Analyze the concavity and the critical points of $f(x)$

Since $(x - 1)^2 \ge 0$ for all real $x$, we have:
$(x - 1)^2 + 1 \ge 1 > 0 \implies f''(x) > 0 \quad \forall x \in \mathbb{R}$

Because $f''(x) > 0$ for all $x \in \mathbb{R}$:

1. The function $f(x)$ is strictly convex on $\mathbb{R}$.
2. The derivative $f'(x)$ is strictly increasing on $\mathbb{R}$.

Since $f'(x)$ is a continuous, strictly increasing function with $\lim_{x \to -\infty} f'(x) = -\infty$ and $\lim_{x \to \infty} f'(x) = +\infty$, by the Intermediate Value Theorem, $f'(x) = 0$ has exactly one real root, which we will call $x = x_0$.

Step 3: Determine the number of real roots of $f(x) = 0$

Since $x_0$ is the unique real root of $f'(x) = 0$:

• $f'(x) < 0$ for $x \in (-\infty, x_0)$, so $f(x)$ is strictly decreasing on $(-\infty, x_0]$.
• $f'(x) > 0$ for $x \in (x_0, \infty)$, so $f(x)$ is strictly increasing on $[x_0, \infty)$.

Thus, $f(x)$ has a unique global minimum at $x = x_0$.

Evaluating $f(x)$ at $x = 0$:
$f(0) = 0^4 - 4(0)^3 + 12(0)^2 + 0 - 1 = -1 < 0$

Since $f(x_0)$ is the global minimum of $f(x)$, we must have:
$f(x_0) \le f(0) = -1 < 0$

Furthermore, observing the limits of $f(x)$ as $x \to \pm\infty$:
$\lim_{x \to -\infty} f(x) = +\infty \quad \text{and} \quad \lim_{x \to \infty} f(x) = +\infty$

By the Intermediate Value Theorem and strict monotonicity:

1. $f(x)$ changes sign from $+\infty$ to $f(x_0) < 0$ on the interval $(-\infty, x_0)$, so $f(x) = 0$ has exactly one real root in $(-\infty, x_0)$.
2. $f(x)$ changes sign from $f(x_0) < 0$ to $+\infty$ on the interval $(x_0, \infty)$, so $f(x) = 0$ has exactly one real root in $(x_0, \infty)$.

Conclusion

The polynomial equation $x^4 - 4x^3 + 12x^2 + x - 1 = 0$ has 2 distinct real roots.