1. Find the Roots and Polynomial Coefficients
- Given roots are $\alpha = 1$ and $\beta = 1 + i\sqrt{2}$.
- Since the polynomial coefficients ($a, b, c$) are real numbers ($a, b, c \in \mathbb{R}$), complex roots must occur in conjugate pairs.
- Therefore, the conjugate of $\beta$, which is $\bar{\beta} = 1 - i\sqrt{2}$, must be the third root ($\gamma$).
- The three roots of $x^3 + ax^2 + bx + c = 0$ are $1$, $1 + i\sqrt{2}$, and $1 - i\sqrt{2}$.
- Using Vieta's formulas:
- Sum of roots: $\alpha + \beta + \gamma = 1 + (1 + i\sqrt{2}) + (1 - i\sqrt{2}) = 3$. This equals $-a/1$, so $a = -3$.
- Sum of products of roots taken two at a time: $\alpha\beta + \alpha\gamma + \beta\gamma = 1(1 + i\sqrt{2}) + 1(1 - i\sqrt{2}) + (1 + i\sqrt{2})(1 - i\sqrt{2}) = 1 + i\sqrt{2} + 1 - i\sqrt{2} + (1 - (i^2 \times 2)) = 2 + (1 - (-2)) = 5$. This equals $b/1$, so $b = 5$.
- Product of roots: $\alpha\beta\gamma = 1 \times (1 + i\sqrt{2}) \times (1 - i\sqrt{2}) = 1 \times (1 - (-2)) = 3$. This equals $-c/1$, so $c = -3$.
- The polynomial is $P(x) = x^3 - 3x^2 + 5x - 3$.
2. Evaluate the Definite Integral
- We need to calculate $\int_{-1}^{1} (x^3 + ax^2 + bx + c) dx$.
- Substitute the values of $a$ and $c$: $\int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) dx$.
- We can split the integral: $\int_{-1}^{1} x^3 dx + \int_{-1}^{1} (-3x^2) dx + \int_{-1}^{1} 5x dx + \int_{-1}^{1} (-3) dx$.
- Use the property that the integral of an odd function over a symmetric interval $[-k, k]$ is zero. $x^3$ and $5x$ are odd functions.
- $\int_{-1}^{1} x^3 dx = 0$
- $\int_{-1}^{1} 5x dx = 0$
- The integral simplifies to $\int_{-1}^{1} (-3x^2 - 3) dx$.
- Evaluate the remaining terms:
- $\int_{-1}^{1} (-3x^2) dx = [-x^3]_{-1}^{1} = -(1)^3 - (-(-1)^3) = -1 - (1) = -2$.
- $\int_{-1}^{1} (-3) dx = [-3x]_{-1}^{1} = (-3 \times 1) - (-3 \times -1) = -3 - 3 = -6$.
- The total value of the integral is $-2 + (-6) = -8$.
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