Let $p(x)$ be a quadratic polynomial such that $p(1) = p(-1) = 0$. What is the coefficient of $x$ in $p(x)$?
Show 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.
Step 1: Understanding the Question:
We are given a quadratic polynomial $p(x)$ that satisfies the conditions $p(1) = 0$ and $p(-1) = 0$.
This means that $x = 1$ and $x = -1$ are the roots (or zeros) of the quadratic polynomial $p(x)$.
We need to determine the coefficient of the linear term, i.e., the coefficient of $x$, in the expansion of $p(x)$. Step 2: Key Formula or Approach:
Any quadratic polynomial with roots $\alpha$ and $\beta$ can be written in the form:
\[ p(x) = a(x - \alpha)(x - \beta) \]
where $a$ is a non-zero real constant.
Alternatively, we can express the polynomial in its standard form:
\[ p(x) = ax^2 + bx + c \]
and apply the given conditions to find the value of $b$. Step 3: Detailed Explanation:
• Let us assume the standard form of the quadratic polynomial is:
\[ p(x) = ax^2 + bx + c \quad (\text{where } a \neq 0) \]
• We substitute the first given condition, $p(1) = 0$, into the standard form:
\[ p(1) = a(1)^2 + b(1) + c = 0 \]
\[ a + b + c = 0 \quad \text{--- (Equation 1)} \]
• We substitute the second given condition, $p(-1) = 0$, into the standard form:
\[ p(-1) = a(-1)^2 + b(-1) + c = 0 \]
\[ a - b + c = 0 \quad \text{--- (Equation 2)} \]
• To solve for the coefficient of $x$ (which is $b$), we subtract Equation 2 from Equation 1:
\[ (a + b + c) - (a - b + c) = 0 - 0 \]
\[ 2b = 0 \]
\[ b = 0 \]
• Alternatively, using the factored form with roots $\alpha = 1$ and $\beta = -1$:
\[ p(x) = a(x - 1)(x - (-1)) \]
\[ p(x) = a(x - 1)(x + 1) \]
\[ p(x) = a(x^2 - 1) \]
\[ p(x) = ax^2 + 0x - a \]
Comparing this with the standard form, the coefficient of $x$ is clearly $0$. Step 4: Final Answer:
The coefficient of $x$ in the quadratic polynomial $p(x)$ is $0$.
