Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 7"

m (→‎Solution)
(→‎Solution)
Line 8: Line 8:
  
 
== Solution ==
 
== Solution ==
−
<math>2x+8</math>
+
Since we're being asked to find a remainder when a polynomial is divided by a quadratic, we can assume that the remainder will be at most linear. Thus, the remainder can be written in the form <math>ax + b</math>.
 +
 
 +
It is given that the polynomial <math>P(x)</math> has a remainder of <math>4</math> when divided by <math>x+2</math> and a remainder of <math>14</math> when divided by <math>x-3</math>, which translates to <math>ax+b = y(x+2) + 4</math> and <math>ax+b = y(x-3) + 14</math>. However, for both of these equations to always be true, the coefficient <math>y</math> must be equal to <math>a</math>.
 +
 
 +
Thus, <math>ax+b = ax+2a+4</math> and <math>ax+b = ax-3a+14</math>.
 +
These equations simplify to <math>b = 2a+4</math> and <math>b = -3a+14</math>, which shows that
 +
<math>2a+4 = -3a+14</math>, so
 +
<math>5a = 10</math>,
 +
meaning <math>a = 2</math>.
 +
 
 +
Plugging <math>a = 2</math> back into either equation gives <math>b = 8</math>, meaning the remainder is <math>2x+8</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:25, 23 December 2019

Problem

A polynomial $P(x)$ has a remainder of $4$ when divided by $x+2$ and a remainder of $14$ when divided by $x-3.$ What is the remainder when $P(x)$ is divided by $(x+2)(x-3)$?


Solution

Since we're being asked to find a remainder when a polynomial is divided by a quadratic, we can assume that the remainder will be at most linear. Thus, the remainder can be written in the form $ax + b$.

It is given that the polynomial $P(x)$ has a remainder of $4$ when divided by $x+2$ and a remainder of $14$ when divided by $x-3$, which translates to $ax+b = y(x+2) + 4$ and $ax+b = y(x-3) + 14$. However, for both of these equations to always be true, the coefficient $y$ must be equal to $a$.

Thus, $ax+b = ax+2a+4$ and $ax+b = ax-3a+14$. These equations simplify to $b = 2a+4$ and $b = -3a+14$, which shows that $2a+4 = -3a+14$, so $5a = 10$, meaning $a = 2$.

Plugging $a = 2$ back into either equation gives $b = 8$, meaning the remainder is $2x+8$.

See also

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_7&oldid=113233"