To find the number of distinct real roots of the equation ${x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$, we can analyze the behavior of the function $P(x) = {x^4} - 4{x^3} + 12{x^2} + x - 1$ using calculus.

Step 1: Find the first derivative of P(x)

To find the critical points (local maxima and minima) of the function, we compute its first derivative: $P'(x) = \frac{d}{dx}({x^4} - 4{x^3} + 12{x^2} + x - 1)$ $P'(x) = 4{x^3} - 12{x^2} + 24x + 1$

Step 2: Analyze the critical points by examining the second derivative

To understand the nature of the first derivative $P'(x)$, we compute the second derivative, $P''(x)$: $P''(x) = \frac{d}{dx}(4{x^3} - 12{x^2} + 24x + 1)$ $P''(x) = 12{x^2} - 24x + 24$ We can factor out 12: $P''(x) = 12({x^2} - 2x + 2)$ To determine the sign of $P''(x)$, we can complete the square for the quadratic term: ${x^2} - 2x + 2 = ({x^2} - 2x + 1) + 1 = (x - 1)^2 + 1$ Since $(x - 1)^2 \ge 0$ for all real $x$, we have $(x - 1)^2 + 1 \ge 1$. Therefore, $P''(x) = 12((x - 1)^2 + 1)$ is always positive for all real values of $x$.

Step 3: Analyze the behavior of P'(x) and find the number of critical points of P(x)

Since $P''(x) > 0$ for all $x$, the first derivative $P'(x)$ is a strictly increasing function. A strictly increasing function can cross the x-axis at most once. This means the equation $P'(x) = 0$ has exactly one real root. Let's locate this root by checking the sign of $P'(x)$ at some points: $P'(0) = 4(0)^3 - 12(0)^2 + 24(0) + 1 = 1 > 0$ $P'(-1) = 4(-1)^3 - 12(-1)^2 + 24(-1) + 1 = -4 - 12 - 24 + 1 = -39 < 0$ Since $P'(x)$ is continuous and changes sign between $x=-1$ and $x=0$, the unique real root of $P'(x)=0$, let's call it $\alpha$, must lie in the interval $(-1, 0)$.

This single root $\alpha$ corresponds to the only critical point of the function $P(x)$. Since $P''(\alpha) > 0$, this critical point is a local minimum. Because it's the only critical point, it is the global minimum of the function.

Step 4: Analyze the behavior of P(x) and determine the number of real roots

The function $P(x)$ is a polynomial of degree 4 with a positive leading coefficient, so: $\lim_{x \to \infty} P(x) = +\infty$ $\lim_{x \to -\infty} P(x) = +\infty$ The function decreases for $x < \alpha$ and increases for $x > \alpha$. It has a global minimum at $x = \alpha$.

To find the number of roots of $P(x)=0$, we need to determine the sign of the minimum value, $P(\alpha)$. We know that $\alpha \in (-1, 0)$. Since $P(x)$ is increasing for $x > \alpha$, we can say that $P(\alpha) < P(0)$. Let's calculate $P(0)$: $P(0) = (0)^4 - 4(0)^3 + 12(0)^2 + (0) - 1 = -1$ So, the minimum value $P(\alpha) < P(0) = -1$. This means the global minimum of the function is negative.

Step 5: Conclusion

Let's summarize the behavior of $P(x)$:

1. As $x \to -\infty$, $P(x) \to +\infty$.
2. The function decreases to a negative minimum value $P(\alpha) < -1$.
3. As $x \to +\infty$, $P(x) \to +\infty$.

Since $P(x)$ is a continuous function, by the Intermediate Value Theorem:

• As $P(x)$ goes from positive infinity down to a negative minimum, it must cross the x-axis exactly once. This gives one real root for $x < \alpha$.
• As $P(x)$ goes from its negative minimum up to positive infinity, it must cross the x-axis exactly once again. This gives a second real root for $x > \alpha$.

Therefore, the equation ${x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$ has exactly two distinct real roots.