Key Concepts and Formulas

• Quadratic Polynomial: A polynomial of degree 2, generally expressed as $p(x) = ax^2 + bx + c$, where $a, b, c$ are constants and $a \neq 0$.
• Remainder Theorem: If a polynomial $p(x)$ is divided by $(x - k)$, the remainder is $p(k)$.
• System of Linear Equations: A set of two or more linear equations containing two or more variables.

Step-by-Step Solution

Step 1: Determine the Constant Term ($c$)

We are given that $p(0) = 1$. This means when we substitute $x = 0$ into the quadratic equation, the result is 1. We can use this to find the value of $c$. Since $p(x) = ax^2 + bx + c$, we have: $p(0) = a(0)^2 + b(0) + c = 1$ $0 + 0 + c = 1$ $c = 1$ Now we know that $p(x) = ax^2 + bx + 1$.

Step 2: Apply the Remainder Theorem for Division by $(x - 1)$

We are given that $p(x)$ leaves a remainder of 4 when divided by $(x - 1)$. By the Remainder Theorem, this means $p(1) = 4$. Substituting $x = 1$ into our expression for $p(x)$, we get: $p(1) = a(1)^2 + b(1) + 1 = 4$ $a + b + 1 = 4$ $a + b = 3$ This gives us our first equation relating $a$ and $b$: $a + b = 3$.

Step 3: Apply the Remainder Theorem for Division by $(x + 1)$

We are given that $p(x)$ leaves a remainder of 6 when divided by $(x + 1)$. By the Remainder Theorem, this means $p(-1) = 6$. Substituting $x = -1$ into our expression for $p(x)$, we get: $p(-1) = a(-1)^2 + b(-1) + 1 = 6$ $a - b + 1 = 6$ $a - b = 5$ This gives us our second equation relating $a$ and $b$: $a - b = 5$.

Step 4: Solve the System of Linear Equations

We now have a system of two linear equations: $a + b = 3$ $a - b = 5$ We can solve this system by adding the two equations to eliminate $b$: $(a + b) + (a - b) = 3 + 5$ $2a = 8$ $a = 4$ Now substitute the value of $a$ back into one of the equations to solve for $b$. Using the first equation: $4 + b = 3$ $b = -1$

Step 5: Construct the Complete Quadratic Polynomial

We have found $a = 4$, $b = -1$, and $c = 1$. Therefore, the quadratic polynomial is: $p(x) = 4x^2 - x + 1$

Step 6: Evaluate the Given Options

Now we need to find which of the options is correct by substituting the given values into the polynomial.

• Option (A): $p(2) = 11$ $p(2) = 4(2)^2 - 2 + 1 = 4(4) - 2 + 1 = 16 - 2 + 1 = 15$ Since $p(2) = 15 \neq 11$, option (A) is incorrect.

• Option (B): $p(2) = 19$ Since $p(2) = 15 \neq 19$, option (B) is incorrect.

• Option (C): $p(-2) = 19$ $p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 2 + 1 = 19$ Since $p(-2) = 19$, option (C) is correct.

• Option (D): $p(-2) = 11$ Since $p(-2) = 19 \neq 11$, option (D) is incorrect.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when applying the Remainder Theorem, especially when dividing by $(x + k)$. Remember to substitute $x = -k$.
• Solving Linear Equations: Double-check your algebra when solving the system of equations to avoid errors in calculating $a$ and $b$.

Summary

By applying the Remainder Theorem and solving the system of linear equations, we found the quadratic polynomial to be $p(x) = 4x^2 - x + 1$. Evaluating the given options, we found that $p(-2) = 19$, which corresponds to option (C).

The final answer is \boxed{19}, which corresponds to option (C).