Key Concepts and Formulas

• Remainder Theorem: For a polynomial $P(x)$, the remainder on division by $(x-a)$ is $P(a)$.
• Definite Integration of Polynomials: The integral of a polynomial $ax^n$ is $\frac{ax^{n+1}}{n+1}$. For a definite integral $\int_a^b f(x) dx$, we evaluate the antiderivative $F(x)$ at the limits: $F(b) - F(a)$.
• Solving Systems of Linear Equations: We will use algebraic manipulation (substitution or elimination) to find the values of unknown coefficients.

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Step-by-Step Solution

Step 1: Understand the Given Information and Goal We are given a quadratic polynomial $P(x) = x^2 + bx + c$ with real coefficients $b$ and $c$. We are provided with two conditions:

1. The definite integral of $P(x)$ from 0 to 1 is 1: $\int_0^1 P(x) dx = 1$.
2. When $P(x)$ is divided by $(x-2)$, the remainder is 5. Our objective is to find the value of $9(b+c)$.

Step 2: Apply the Remainder Theorem The second condition states that $P(x)$ leaves a remainder of 5 when divided by $(x-2)$. By the Remainder Theorem, this means $P(2) = 5$. Substituting $x=2$ into the polynomial $P(x) = x^2 + bx + c$: $P(2) = (2)^2 + b(2) + c$ $P(2) = 4 + 2b + c$ Since $P(2) = 5$, we have: $4 + 2b + c = 5$ Subtracting 4 from both sides gives us our first linear equation: $2b + c = 1 \quad (\text{Equation 1})$ Reasoning: This step converts the information about polynomial division into a concrete algebraic equation involving the unknown coefficients $b$ and $c$.

Step 3: Apply the Definite Integration Condition The first condition states that $\int_0^1 P(x) dx = 1$. Substituting $P(x) = x^2 + bx + c$ into the integral: $\int_0^1 (x^2 + bx + c) dx = 1$ Now, we find the antiderivative of $P(x)$ using the power rule for integration: $\left[ \frac{x^{2+1}}{2+1} + \frac{bx^{1+1}}{1+1} + \frac{cx^{0+1}}{0+1} \right]_0^1 = 1$ $\left[ \frac{x^3}{3} + \frac{bx^2}{2} + cx \right]_0^1 = 1$ Next, we evaluate this antiderivative at the upper limit ($x=1$) and subtract its value at the lower limit ($x=0$), according to the Fundamental Theorem of Calculus: $\left( \frac{1^3}{3} + \frac{b(1)^2}{2} + c(1) \right) - \left( \frac{0^3}{3} + \frac{b(0)^2}{2} + c(0) \right) = 1$ $\left( \frac{1}{3} + \frac{b}{2} + c \right) - (0) = 1$ $\frac{1}{3} + \frac{b}{2} + c = 1$ To isolate the terms with $b$ and $c$, subtract $\frac{1}{3}$ from both sides: $\frac{b}{2} + c = 1 - \frac{1}{3}$ $\frac{b}{2} + c = \frac{2}{3} \quad (\text{Equation 2})$ Reasoning: This step utilizes calculus to establish a second independent relationship between $b$ and $c$, derived from the given integral condition.

Step 4: Solve the System of Linear Equations We now have a system of two linear equations with two variables:

1. $2b + c = 1$
2. $\frac{b}{2} + c = \frac{2}{3}$

We can solve this system. Let's use the substitution method. From Equation 1, we can express $c$ in terms of $b$: $c = 1 - 2b$ Substitute this expression for $c$ into Equation 2: $\frac{b}{2} + (1 - 2b) = \frac{2}{3}$ Combine the terms involving $b$: $\frac{b}{2} - 2b + 1 = \frac{2}{3}$ To combine $\frac{b}{2}$ and $-2b$, find a common denominator: $\frac{b}{2} - \frac{4b}{2} + 1 = \frac{2}{3}$ $-\frac{3b}{2} + 1 = \frac{2}{3}$ Subtract 1 from both sides: $-\frac{3b}{2} = \frac{2}{3} - 1$ $-\frac{3b}{2} = \frac{2}{3} - \frac{3}{3}$ $-\frac{3b}{2} = -\frac{1}{3}$ Multiply both sides by $-1$: $\frac{3b}{2} = \frac{1}{3}$ To solve for $b$, multiply both sides by $\frac{2}{3}$: $b = \frac{1}{3} \times \frac{2}{3}$ $b = \frac{2}{9}$ Now, substitute the value of $b$ back into the expression for $c$ ($c = 1 - 2b$): $c = 1 - 2\left(\frac{2}{9}\right)$ $c = 1 - \frac{4}{9}$ $c = \frac{9}{9} - \frac{4}{9}$ $c = \frac{5}{9}$ Reasoning: By systematically solving the system of equations, we determine the unique values of $b$ and $c$ that satisfy both given conditions.

Step 5: Calculate the Final Value We are asked to find the value of $9(b+c)$. Substitute the values of $b$ and $c$ we found: $b+c = \frac{2}{9} + \frac{5}{9}$ $b+c = \frac{2+5}{9}$ $b+c = \frac{7}{9}$ Now, multiply this sum by 9: $9(b+c) = 9 \times \left(\frac{7}{9}\right)$ $9(b+c) = 7$ Reasoning: This final step directly addresses the question asked in the problem by combining the determined coefficients and performing the required multiplication.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when manipulating fractions and solving the system of equations. Double-checking each step can prevent small errors from snowballing.
• Integration Errors: Ensure the power rule for integration is applied correctly, especially with the addition of the constant of integration (though it cancels out in definite integrals).
• Remainder Theorem Misapplication: Always ensure you are substituting the correct value of $x$ into $P(x)$ based on the divisor $(x-a)$.

Summary The problem required us to find the value of $9(b+c)$ for a quadratic polynomial $P(x) = x^2 + bx + c$. We used the Remainder Theorem to derive one linear equation from the division condition ($P(2)=5$). We then used the definite integration condition ($\int_0^1 P(x) dx = 1$) to derive a second linear equation. Solving these two equations simultaneously yielded the values of $b$ and $c$. Finally, we substituted these values into the expression $9(b+c)$ to obtain the answer.

The final answer is $\boxed{7}$.