Let P(x) = x^2 + bx + c be a quadratic polyno | vedclass

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(C) Given $P(x) = x^2 + bx + c$.Integrating $P(x)$ from $0$ to $1$:$\int_{0}^{1} (x^2 + bx + c) dx = [\frac{x^3}{3} + \frac{bx^2}{2} + cx]_{0}^{1} = \

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$7$

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(C) Given $P(x) = x^2 + bx + c$.Integrating $P(x)$ from $0$ to $1$:$\int_{0}^{1} (x^2 + bx + c) dx = [\frac{x^3}{3} + \frac{bx^2}{2} + cx]_{0}^{1} = \frac{1}{3} + \frac{b}{2} + c = 1$.Multiplying by $6$,we get $2 + 3b + 6c = 6$,so $3b + 6c = 4$ ... $(1)$.By the Remainder Theorem,$P(2) = 5$:$2^2 + b(2) + c = 5 \Rightarrow 4 + 2b + c = 5 \Rightarrow 2b + c = 1$ ... $(2)$.From $(2)$,$c = 1 - 2b$. Substituting into $(1)$:$3b + 6(1 - 2b) = 4 \Rightarrow 3b + 6 - 12b = 4 \Rightarrow -9b = -2 \Rightarrow b = \frac{2}{9}$.Then $c = 1 - 2(\frac{2}{9}) = 1 - \frac{4}{9} = \frac{5}{9}$.Thus,$b + c = \frac{2}{9} + \frac{5}{9} = \frac{7}{9}$.Therefore,$9(b + c) = 9(\frac{7}{9}) = 7$.

Let $P(x) = x^2 + bx + c$ be a quadratic polynomial with real coefficients such that $\int_{0}^{1} P(x) dx = 1$ and $P(x)$ leaves a remainder of $5$ when divided by $(x-2)$. Then the value of $9(b+c)$ is equal to:

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