In the open interval (0, 1), the polynomial p(x) = x⁴ - 4x³ + 2 has

To determine the number of real roots of the polynomial $p\left(x\right)=x^4-4x^3+2$ in the open interval (0, 1), we can analyze the behavior of the function and its values at the boundary points of the interval.

First, let us evaluate the polynomial at the endpoints of the interval, $x=0$ and $x=1$:

At $x=0$:
$p\left(0\right)=0^4-4\left(0^3\right)+2=2>0$

At $x=1$:
$p\left(1\right)=1^4-4\left(1^3\right)+2=1-4+2=-1<0$

Since the polynomial $p\left(x\right)$ is continuous on the closed interval [0, 1], and the values at the endpoints have opposite signs ($p\left(0\right)>0$ and $p\left(1\right)<0$), the Intermediate Value Theorem guarantees that there is at least one real root in the open interval (0, 1).

To determine if there is more than one root, we examine the derivative of $p\left(x\right)$ to understand its monotonicity in the interval (0, 1):

$p^{\prime }\left(x\right)=4x^3-12x^2=4x^2(x-3)$

For any $x$ in the open interval (0, 1):
- The term $4x^2$ is always positive ($4x^2>0$).
- The term $x-3$ is always negative because $x<1$, which implies $x-3<-2<0$.

Consequently, the product $p^{\prime }\left(x\right)=4x^2(x-3)<0$ for all $x$ in (0, 1).

Since the derivative is strictly negative on the interval (0, 1), the function $p\left(x\right)$ is strictly decreasing on this interval. A strictly monotonic function can cross the x-axis at most once. Therefore, there is exactly one real root in the open interval (0, 1).