# 1965 AHSME Problems/Problem 22

## Problem

If $a_2 \neq 0$ and $r$ and $s$ are the roots of $a_0 + a_1x + a_2x^2 = 0$, then the equality $a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )$ holds:

$\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 \qquad \textbf{(B) }\ \text{for all values of }x \\ \textbf{(C) }\ \text{only when }x = 0 \qquad \textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0$

## Solution

$a_0$ cannot be $0$, because $a_2 \neq 0$, so the polynomial can take non-zero values when $a_0=0$ and thereby not satisfy the equation. Expanding $a_0(1-\frac{x}{r})(1-\frac{x}{s})$ gives us $a_0(1-\frac{x}{r}-\frac{x}{s}+\frac{x^2}{rs})$. Using Vieta's formulas, we see that $rs=\frac{a_0}{a_2}$, so $a_0=a_2rs$. Plugging this into the expanded right hand side of the given equation, we see that that side equals $a_0-a_2(r+s)x+a_2x^2$. Using Vieta's formulas again, we equate this expression to $a_0+a_1x+a_2x^2$, which is the left hand side of the given equation. Thus, as long as $a_0 \neq 0$, the equation holds for all values of $x$. This fact corresponds with answer choice $\fbox{\textbf{(A)}}$.