1964 AHSME Problems/Problem 25
Problem 25
The set of values of for which has two factors, with integer coefficients, which are linear in and , is precisely:
Solution
Since there are only candidate values for , we test .
If , then the expression is . The term appears in each monomial, giving , which has two factors that are linear in and with integer coefficients. This eliminates and .
We next test . This gives . The linear factors will be of the form . We will match coefficients from the skeleton form to the expression that is to be factored.
Matching the coefficient gives , and WLOG we can let to give .
Matching the coefficients gives . Since are integers, we either have . WLOG we can pick the former, since if is a factorization, so is .
We now have . Matching the term gives . We now have . Matching the term gives , which leads to . Finally, matching constants leads to and for a provisional factorization of . We've matched every coefficient except for , and finding the term by selectively multiplying leads to , which matches, so this is a real factorization.
We now must check . The expression is , and the skeleton once again is . The beginning three steps matching are the same, leading to . Matching the term gives , or . The skeleton becomes . The constant locks in for a provisional factorization of . However, the term does not match up, because it is , when it needs to be .
Thus, work and does not, giving answer .
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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