1. **State the problem:** We have the function $$f(x) = x^4 - ax^3 + ax^2 - 2x + 1$$ with $$a = 2$$ and the factorization $$f(x) = (x - 1) \cdot g(x)$$. We want to find the values of $$x \in \mathbb{R}$$ such that $$f(x) \geq 0$$. 2. **Substitute the value of $$a$$:** $$f(x) = x^4 - 2x^3 + 2x^2 - 2x + 1$$ 3. **Use the factorization:** Since $$f(x) = (x - 1) \cdot g(x)$$, divide $$f(x)$$ by $$x - 1$$ to find $$g(x)$$. 4. **Polynomial division:** Divide $$x^4 - 2x^3 + 2x^2 - 2x + 1$$ by $$x - 1$$: - Leading term: $$x^4 / x = x^3$$ - Multiply: $$x^3(x - 1) = x^4 - x^3$$ - Subtract: $$(-2x^3) - (-x^3) = -x^3$$ - Bring down $$+ 2x^2$$ - Next term: $$-x^3 / x = -x^2$$ - Multiply: $$-x^2(x - 1) = -x^3 + x^2$$ - Subtract: $$2x^2 - x^2 = x^2$$ - Bring down $$- 2x$$ - Next term: $$x^2 / x = x$$ - Multiply: $$x(x - 1) = x^2 - x$$ - Subtract: $$-2x - (-x) = -x$$ - Bring down $$+ 1$$ - Next term: $$-x / x = -1$$ - Multiply: $$-1(x - 1) = -x + 1$$ - Subtract: $$1 - 1 = 0$$ So, $$g(x) = x^3 - x^2 + x - 1$$. 5. **Rewrite $$f(x)$$:** $$f(x) = (x - 1)(x^3 - x^2 + x - 1)$$ 6. **Factor $$g(x)$$ further:** Try to factor $$g(x) = x^3 - x^2 + x - 1$$ by grouping: $$x^3 - x^2 + x - 1 = (x^3 - x^2) + (x - 1) = x^2(x - 1) + 1(x - 1) = (x - 1)(x^2 + 1)$$ 7. **Complete factorization:** $$f(x) = (x - 1)^2 (x^2 + 1)$$ 8. **Analyze the inequality $$f(x) \geq 0$$:** - Since $$x^2 + 1 > 0$$ for all real $$x$$, - The sign of $$f(x)$$ depends on $$(x - 1)^2$$ which is always $$\geq 0$$. 9. **Conclusion:** $$f(x) \geq 0$$ for all real $$x$$, and $$f(x) = 0$$ only at $$x = 1$$. **Final answer:** $$\boxed{\text{For all } x \in \mathbb{R}, f(x) \geq 0 \text{ and } f(x) = 0 \text{ only at } x = 1}$$