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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 \( \boxed{2} \).

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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.

The polynomial x^3 + px^2 - x - 2, p ∈ ℝ, has x - 1 as a factor. What is the value of p?

Explore how to solve the polynomial equation x^3 + px^2 - x - 2 when x - 1 is a factor, and learn how to find the value of p. This problem covers important algebraic concepts such as the Factor Theorem and polynomial roots. Suitable for Grades 10-12.

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