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

The correct option is: One real root

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 use the Intermediate Value Theorem and analyze the behavior of the derivative of the function.

Step 1: Evaluate the function at the endpoints of the interval

Let us calculate the values of the polynomial at the boundaries of the interval, namely at

$x=0$

and

$x=1$

At the left endpoint:

$p\left(0\right)=0^4-4\left(0^3\right)+2=2$

Since this value is positive:

$p\left(0\right)>0$

At the right endpoint:

$p\left(1\right)=1^4-4\left(1^3\right)+2=1-4+2=-1$

Since this value is negative:

$p\left(1\right)<0$

Since the polynomial is a continuous function on the closed interval [0, 1] and it changes sign from positive to negative, the Intermediate Value Theorem guarantees that there exists at least one real root in the open interval (0, 1).

Step 2: Determine if the root is unique using the derivative

To find out if there can be more than one root, we differentiate the function to check its monotonicity:

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

We can factor the derivative as:

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

Now, let us analyze the sign of the derivative in the open interval (0, 1):

1. The term

$4x^2$

is strictly positive for all values of x in the interval (0, 1).

2. The term

$(x-3)$

is strictly negative for all values of x in the interval (0, 1) because x is strictly less than 1, meaning x - 3 is less than -2.

Since the product of a positive number and a negative number is always negative, we have:

$p^{\text{'}}\left(x\right)<0$

for all x in (0, 1). This means that the function is strictly decreasing on the interval.

Conclusion

A strictly decreasing function can cross the x-axis at most once. Since the Intermediate Value Theorem established that it crosses at least once, the polynomial has exactly one real root in the open interval (0, 1).