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Difference between revisions of "1955 AHSME Problems/Problem 25"
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<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>.
  
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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>.
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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>
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==Solution 2 (direct factorization)==
 
==Solution 2 (direct factorization)==
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</cmath>
 
</cmath>
 
where both quadratics are irreducible (over the field of real numbers).
 
where both quadratics are irreducible (over the field of real numbers).
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Hence none of the given options is a factor. So the answer is <math>\boxed{(\textbf{C})}</math>
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Hence none of the given options is a factor. So the answer is <math>\boxed{(\textbf{E})}</math>
  
 
~VensL
 
~VensL
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==See Also==
 
==See Also==
  

Latest revision as of 17:42, 21 June 2022

Problem 25

One of the factors of $x^4+2x^2+9$ is:

$\textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these}$

Solution

We can test each of the answer choices by using polynomial division.

$x^2 + 3$ leaves behind a remainder, and so does $x^2 - 3$.

In addition, $x + 1$ also fails the test, and that takes down $x^2 - 2x - 3$, which can be expressed as $(x + 1)(x - 3)$. That leaves $\boxed{(\textbf{E})}$

Solution 2 (direct factorization)

Notice the leading and constant terms are begging us to create a binomial. So \[x^4 + 2x^2 + 9 = (x^4 + 6x^2 + 9) - 4x^2 = (x^2 + 3)^2 - (2x)^2 = (x^2 + 2x + 3)(x^2 - 2x + 3),\] where both quadratics are irreducible (over the field of real numbers). Hence none of the given options is a factor. So the answer is $\boxed{(\textbf{E})}$

~VensL

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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All AHSME Problems and Solutions


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