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

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== Problem ==
 
== Problem ==
  
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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
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by <math>x-3</math>. What is the remainder when <math>P(x)</math> is divided by <math>(x+2)(x-3)</math>?
  
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
+
== Solution ==
by <math>x-3.</math> What is the remainder when <math>P(x)</math> is divided by <math>(x+2)(x-3)</math>?
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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>.
  
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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>.
  
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Thus, <math>ax+b = ax+2a+4</math> and <math>ax+b = ax-3a+14</math>.
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These equations simplify to <math>b = 2a+4</math> and <math>b = -3a+14</math>, which shows that
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<math>2a+4 = -3a+14</math>, so
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<math>5a = 10</math>,
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meaning <math>a = 2</math>.
  
== Solution ==
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Plugging <math>a = 2</math> back into either equation gives <math>b = 8</math>, meaning the remainder is <math>2x+8</math>.
{{solution}}
 
  
 
== See also ==
 
== See also ==

Latest revision as of 12:25, 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 (ProblemsAnswer KeyResources)
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
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