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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