Difference between revisions of "2010 AIME II Problems/Problem 6"
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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. | 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. | ||
− | By the [[ | + | 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> | <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> |
Revision as of 05:00, 28 December 2014
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.