Difference between revisions of "1955 AHSME Problems/Problem 25"
Angrybird029 (talk | contribs) (→See Also) |
m (→Solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 9: | Line 9: | ||
<math>x^2 + 3</math> leaves behind a remainder, and so does <math>x^2 - 3</math>. | <math>x^2 + 3</math> leaves behind a remainder, and so does <math>x^2 - 3</math>. | ||
− | In addition, <math>x + 1</math> also fails the test, and that takes down <math>x^2 - 2x - 3</math>, which can be expressed as <math>(x + 1)(x - 3)</math>. | + | In addition, <math>x + 1</math> also fails the test, and that takes down <math>x^2 - 2x - 3</math>, which can be expressed as <math>(x + 1)(x - 3)</math>. That leaves <math>\boxed{(\textbf{E})}</math> |
+ | |||
+ | ==Solution 2 (direct factorization)== | ||
+ | Notice the leading and constant terms are begging us to create a binomial. So | ||
+ | <cmath> | ||
+ | x^4 + 2x^2 + 9 = (x^4 + 6x^2 + 9) - 4x^2 = (x^2 + 3)^2 - (2x)^2 = (x^2 + 2x + 3)(x^2 - 2x + 3), | ||
+ | </cmath> | ||
+ | where both quadratics are irreducible (over the field of real numbers). | ||
+ | Hence none of the given options is a factor. So the answer is <math>\boxed{(\textbf{E})}</math> | ||
+ | |||
+ | ~VensL | ||
− | |||
==See Also== | ==See Also== | ||
Latest revision as of 17:42, 21 June 2022
Problem 25
One of the factors of is:
Solution
We can test each of the answer choices by using polynomial division.
leaves behind a remainder, and so does .
In addition, also fails the test, and that takes down , which can be expressed as . That leaves
Solution 2 (direct factorization)
Notice the leading and constant terms are begging us to create a binomial. So where both quadratics are irreducible (over the field of real numbers). Hence none of the given options is a factor. So the answer is
~VensL
See Also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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.