Nature of factors for x^2 − x + 1 over the reals: determine whether proper linear factors exist.

Introduction / Context:
The factorability of a quadratic over the reals depends on its discriminant. If the discriminant is negative, the quadratic has no real roots and therefore cannot be factored into real linear factors. We analyze x^2 − x + 1 accordingly.

Given Data / Assumptions:

• Quadratic: x^2 − x + 1
• Coefficients are real.

Concept / Approach:
Compute the discriminant D = b^2 − 4ac for a = 1, b = −1, c = 1. Interpret D: D > 0 ⇒ two distinct real linear factors; D = 0 ⇒ a repeated real linear factor; D < 0 ⇒ no real linear factors.

Step-by-Step Solution:
a = 1, b = −1, c = 1D = (−1)^2 − 4*1*1 = 1 − 4 = −3Since D < 0, there are no real roots and no factorization into real linear factors.

Verification / Alternative check:
Graph y = x^2 − x + 1 has vertex above the x-axis and never crosses, confirming no real zeros.

Why Other Options Are Wrong:
“One/two proper linear factor(s)” require D ≥ 0; “exactly one repeated linear factor” corresponds to D = 0, which is not the case; “None of these” is unnecessary.

Common Pitfalls:
Miscomputing D or assuming any quadratic factors over the reals. Always check the discriminant first.

Final Answer:
no proper linear factor