2009 UNCO Math Contest II Problems/Problem 7

Revision as of 12:25, 23 December 2019 by Skyguy88 (talk | contribs) (Solution)

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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