1951 AHSME Problems/Problem 18

Problem

The expression $21x^2 +ax +21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be one if $a$ is:

$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$ $\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$

Solution

We can factor $21x^2 \plus{} ax \plus{} 21$ (Error compiling LaTeX. Unknown error_msg) as $(7x+3)(3x+7)$ , which expands to $21x^2+42x+21$ . So the answer is $\textbf{(D)}\ \text{some even number}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1951 AHSME Problems/Problem 18

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