## What Am I Missing?

This math team problem is giving me a headache. It is one of 20 problems from the 1999 ICTM State Competition for Algebra II. The students have 50 minutes. I’ve spent hours on this one.

Given, $f\left( x \right)=x^{1998}$, $g\left( x \right)=x^{2}-5x+6$, Let $R\left( x \right)$ be the remainder of $\frac{f\left( x \right)}{g\left( x \right)}$, the y-intercept of which can be expressed as $ab\left( a^{c}-b^{c} \right)$. Find $b-a+c$.

I think there must be part of the remainder theorem I’ve either forgotten or never learned. I’ve backed into the solution1, I just don’t know why I can do it.

Ideas?

1The answer is in comments. I don’t want to ruin anyone’s fun.

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### 4 Responses to What Am I Missing?

1. Jackie says:

The answer is 1997, with a = 2, b = 3, and c = 1998.

2. John Armstrong says:

Factor $g(x)=(x-2)(x-3)$. Then we have $f(x) = x^{1998} = Q(x)(x-2)(x-3) + (mx+n)$

because the remainder must be a linear polynomial. Comparing constant terms, we must have $n = -(-2)(-3)k$

as the y-intercept you’re interested in. Any clearer yet?

3. Jackie says:

That’s how I backed into the solution by testing $f\left( x \right)=x^{4}$, $x^{5}$

Thank you.

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