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 $\fbox{\textbf{(A)}}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1965 AHSME Problems/Problem 22

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