Difference between revisions of "1951 AHSME Problems/Problem 18"

Revision as of 20:19, 30 April 2015

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 $21x^2 + ax + 21$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 == Solution ==
-We can factor <math> 21x^2 \plus{} ax \plus{} 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
+We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
 == See Also ==

Page

Toolbox

Search

Difference between revisions of "1951 AHSME Problems/Problem 18"

Revision as of 20:19, 30 April 2015

Problem

Solution

See Also