Key Concepts and Formulas

• A quadratic polynomial with roots $r_1$ and $r_2$ can be expressed as $f(x) = a(x - r_1)(x - r_2)$, where $a$ is a non-zero constant.
• Given $f(x) = ax^2 + bx + c$, the sum of the roots is $-b/a$, and the product of the roots is $c/a$.
• Interval notation: $(a, b)$ denotes the open interval between $a$ and $b$, excluding $a$ and $b$.

Step-by-Step Solution

1. Define the Quadratic Polynomial

We are given that $f(x)$ is a quadratic polynomial and one of its roots is 3. Let the other root be $\alpha$. We can express $f(x)$ in terms of its roots as: $f(x) = a(x - 3)(x - \alpha)$ where $a$ is a non-zero constant.

Explanation: This is the standard form for a quadratic polynomial given its roots. The leading coefficient 'a' is necessary to fully define the quadratic.

2. Evaluate the Polynomial at x = -1 and x = 2

We are given that $f(-1) + f(2) = 0$. Let's find expressions for $f(-1)$ and $f(2)$ using the polynomial defined in Step 1.

For $x = -1$: $f(-1) = a(-1 - 3)(-1 - \alpha)$ $f(-1) = a(-4)(-1 - \alpha)$ $f(-1) = 4a(1 + \alpha)$

For $x = 2$: $f(2) = a(2 - 3)(2 - \alpha)$ $f(2) = a(-1)(2 - \alpha)$ $f(2) = -a(2 - \alpha)$ $f(2) = a(\alpha - 2)$

Explanation: Substituting $x = -1$ and $x = 2$ into the expression for $f(x)$ allows us to use the given condition to solve for the unknown root $\alpha$.

3. Apply the Given Condition f(-1) + f(2) = 0

Substitute the expressions for $f(-1)$ and $f(2)$ into the given condition: $4a(1 + \alpha) + a(\alpha - 2) = 0$

Since $a \neq 0$, we can divide the entire equation by $a$: $4(1 + \alpha) + (\alpha - 2) = 0$

Expand and simplify: $4 + 4\alpha + \alpha - 2 = 0$ $5\alpha + 2 = 0$ $5\alpha = -2$ $\alpha = -\frac{2}{5}$

Explanation: The condition $f(-1) + f(2) = 0$ provides a relationship between the roots of the polynomial. Dividing by 'a' (since it's nonzero) simplifies the equation, allowing us to solve for $\alpha$.

4. Determine the Interval for the Other Root

We found that the other root is $\alpha = -\frac{2}{5} = -0.4$. Now, we need to determine which of the given intervals contains this value.

• Option (A): $(-3, -1)$. Since $-3 < -0.4$ is FALSE, $-0.4$ is not in this interval.
• Option (B): $(1, 3)$. Since $1 < -0.4$ is FALSE, $-0.4$ is not in this interval.
• Option (C): $(-1, 0)$. Since $-1 < -0.4 < 0$, $-0.4$ is in this interval.
• Option (D): $(0, 1)$. Since $0 < -0.4$ is FALSE, $-0.4$ is not in this interval.

Explanation: We check each interval to see if it contains the calculated root. The interval $(-1, 0)$ is the only one that satisfies the inequality.

Common Mistakes & Tips

• When working with quadratic polynomials and their roots, expressing the polynomial in the form $f(x) = a(x - r_1)(x - r_2)$ is often very helpful.
• Remember that the leading coefficient $a$ of a quadratic polynomial cannot be zero.
• Pay close attention to signs when substituting values into expressions, especially when dealing with negative numbers.

Summary

By expressing the quadratic polynomial in terms of its roots, using the given condition $f(-1) + f(2) = 0$, and solving for the unknown root, we found that the other root is $\alpha = -\frac{2}{5} = -0.4$. This value lies within the interval $(-1, 0)$.

Final Answer The final answer is \boxed{(-1, 0)}, which corresponds to option (C).