Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 7"
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== Problem == | == Problem == | ||
+ | A 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>. What is the remainder when <math>P(x)</math> is divided by <math>(x+2)(x-3)</math>? | ||
− | + | == 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 <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 == |
Latest revision as of 12:25, 23 December 2019
Problem
A polynomial has a remainder of when divided by and a remainder of when divided by . What is the remainder when is divided by ?
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 .
It is given that the polynomial has a remainder of when divided by and a remainder of when divided by , which translates to and . However, for both of these equations to always be true, the coefficient must be equal to .
Thus, and . These equations simplify to and , which shows that , so , meaning .
Plugging back into either equation gives , meaning the remainder is .
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 |