The problem states:

"The polynomial $x^3 + px^2 - x - 2$, with $p \in \mathbb{R}$, has $x - 1$ as a factor. What is the value of $p$?"

Solution:

If $x - 1$ is a factor of the polynomial, then $x = 1$ is a root of the polynomial. That means substituting $x = 1$ into the polynomial should give us zero.

The polynomial is: $f(x) = x^3 + px^2 - x - 2$

Substitute $x = 1$ into this expression: $f(1) = (1)^3 + p(1)^2 - (1) - 2$ $f(1) = 1 + p - 1 - 2 = 0$ Simplifying: $p - 2 = 0$ $p = 2$

Therefore, the value of $p$ is $\boxed{2}$.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you factorize a cubic polynomial when one root is known?
2. What other techniques can be used to find the roots of a cubic polynomial?
3. How can the remainder theorem be applied to check if a factor divides a polynomial?
4. What does it mean when a polynomial has multiple roots?
5. How does synthetic division help in dividing polynomials by linear factors?

Tip: When testing if $x - r$ is a factor of a polynomial, substitute $r$ into the polynomial. If the result is zero, $x - r$ is indeed a factor.