Question

Let $p(x)$ be a quadratic polynomial such that $p(1) = p(-1) = 0$. What is the coefficient of $x$ in $p(x)$?

• A. 0
• B. 1
• C. -1
• D. 2

Hint:

If a polynomial $p(x)$ has roots that are symmetric about the origin (like $\alpha$ and $-\alpha$), then the polynomial is symmetric.
For a quadratic polynomial, this means it is an even function, and thus the coefficient of the odd power of $x$ (which is $x^1$) must be zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The values for which $p(x) = 0$ are the roots of the polynomial.
Key Formula or Approach:
A quadratic with roots $\alpha, \beta$ is \( p(x) = a(x - \alpha)(x - \beta) \).

Step 2: Detailed Explanation:
Roots are $\alpha = 1$ and $\beta = -1$.
\[ p(x) = a(x - 1)(x - (-1)) = a(x - 1)(x + 1) \]
Using the identity $(x-1)(x+1) = x^{2} - 1$:
\[ p(x) = a(x^{2} - 1) = ax^{2} - a \]
In the standard form $ax^{2} + bx + c$, the coefficient of $x$ is $b$.
Here, the polynomial has no $x$ term, so $b = 0$.

Step 3: Final Answer:
The coefficient is 0.
This matches option (A).