Difference between revisions of "2010 AIME II Problems/Problem 6"
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− | ==Problem== | + | == Problem == |
− | + | Find the smallest positive integer <math>n</math> with the property that the [[polynomial]] <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients. | |
− | Find the smallest positive integer n with the property that the polynomial <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients. | ||
==Solution== | ==Solution== | ||
− | + | There are two ways for a monic fourth degree polynomial to be factored into two non-constant polynomials with real coefficients: into a [[cubic]] and a linear equation, or 2 [[quadratic]]s. | |
− | There are | ||
<br/> | <br/> | ||
− | + | *'''Case 1''': The factors are cubic and linear. | |
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− | + | Let <math>x-r_1</math> be the linear root, where <math>r_1</math> is a root of the given quartic, and let <math>x^3+ax^2+bx+c</math> be the cubic. | |
− | <math> | + | By the [[rational roots theorem]], then <math>r_1=1,3,7, 9</math>, or <math>63</math>. Observe that |
− | + | <cmath>(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1.</cmath> | |
− | <math>c-br_1=-n\ | + | Setting [[coefficient]]s equal, we have <math>a-r_1=0 \Longrightarrow a=r_1</math>, <math>b-ar_1=0\Longrightarrow b=a^2</math>, and <math>c-br_1=-n \Longrightarrow n=a^3-c</math>, and <math>-cr_1=63 \Longrightarrow c=\frac{-63}{a}</math>. |
<br/> | <br/> | ||
− | + | It follows that <math>n=a^3+\frac{63}{a}</math>, <math>r_1=1,3,7, 9</math>, or <math>63</math>, which reach minimum when <math>r_1=3</math>, where <math>n=48</math>. | |
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<br/> | <br/> | ||
− | + | *'''Case 2''': The factors are quadratics. | |
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− | + | Let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, so that | |
− | < | + | <cmath>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.</cmath> |
− | and <math>bd = 63</math>. | + | Therefore, again setting coefficients equal, <math>a + c = 0\Longrightarrow a=-c</math>, <math>b + d + ac = 0\Longrightarrow b+d=a^2</math> , <math>ad + bc = - n</math>, and so <math>bd = 63</math>. |
− | <math>b+d=a^2</math> , | + | Since <math>b+d=a^2</math>, the only possible values for <math>(b,d)</math> are <math>(1,63)</math> and <math>(7,9)</math>. From this we find that the possible values for <math>n</math> are <math>\pm 8 \cdot 62</math> and <math>\pm 4 \cdot 2</math>. Therefore, the answer is <math>\boxed{008}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=5|num-a=7|n=II}} | {{AIME box|year=2010|num-b=5|num-a=7|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 11:45, 6 April 2010
Problem
Find the smallest positive integer with the property that the polynomial
can be written as a product of two nonconstant polynomials with integer coefficients.
Solution
There are two ways for a monic fourth degree polynomial to be factored into two non-constant polynomials with real coefficients: into a cubic and a linear equation, or 2 quadratics.
- Case 1: The factors are cubic and linear.
Let be the linear root, where
is a root of the given quartic, and let
be the cubic.
By the rational roots theorem, then , or
. Observe that
Setting coefficients equal, we have ,
, and
, and
.
It follows that ,
, or
, which reach minimum when
, where
.
- Case 2: The factors are quadratics.
Let and
be the two quadratics, so that
Therefore, again setting coefficients equal, ,
,
, and so
.
Since , the only possible values for
are
and
. From this we find that the possible values for
are
and
. Therefore, the answer is
.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |