5. When dividing the polynomial $p$ of degree 2006 by $x-1$, we obtained a remainder of 3, and when dividing it by $x-3$, we obtained a remainder of 5. What remainder do we get when the polynomial $p$ is divided by $(x-1)(x-3)$?

# 50th Mathematical Competition for High School Students in Slovenia

Optional Competition, March 29, 2006

## Problems for 4th Year
III/5. Denimo, da imamo polinom $p(x)$. Če ga delimo z $(x-1)(x-3)$, imamo $p(x)=$ $(x-1)(x-3) \cdot q(x)+r(x)$, kjer je ostanek $r(x)$ stopnje 1 ali manj. Pišimo $r(x)=a x+b$. Ker je ostanek pri deljenju polinoma $p(x)$ z $x-\alpha$ enak $p(\alpha)$, velja:

$$
\begin{aligned}
& 3=p(1)=(1-1)(1-3) \cdot q(1)+a+b=a+b \quad \text { in } \\
& 5=p(3)=(3-1)(3-3) \cdot q(3)+3 a+b=3 a+b
\end{aligned}
$$

Iz prve enakosti dobimo $a=3-b$. To vstavimo v drugo: $5=9-2 b$, od koder izrazimo $b=2$. Končno dobimo še $a=1$. Iskani ostanek je $r(x)=x+2$.

![](https://cdn.mathpix.com/cropped/2024_06_07_fb5898f0472fc84ca82bg-11.jpg?height=60&width=1633&top_left_y=1004&top_left_x=220)

Ugotovitev: stopnja ostanka $r(x)$ je enaka $\mathbf{1}$ ali manj, $r(x)=a x+b \ldots \ldots . .2$ točki

Uporaba dejstva, da je $p(\alpha)$ ostanek pri deljenju polinoma $p(x) \mathbf{z} x-\alpha \ldots \ldots \mathbf{1}$ točka

Nastavitev sistema dveh enačb z dvema neznankama in rešitev le-tega ..... 3 točke

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III/5. Suppose we have a polynomial $p(x)$. If we divide it by $(x-1)(x-3)$, we have $p(x) = (x-1)(x-3) \cdot q(x) + r(x)$, where the remainder $r(x)$ is of degree 1 or less. Let $r(x) = ax + b$. Since the remainder when dividing the polynomial $p(x)$ by $x - \alpha$ is $p(\alpha)$, we have:

$$
\begin{aligned}
& 3 = p(1) = (1-1)(1-3) \cdot q(1) + a + b = a + b \quad \text{and} \\
& 5 = p(3) = (3-1)(3-3) \cdot q(3) + 3a + b = 3a + b
\end{aligned}
$$

From the first equation, we get $a = 3 - b$. Substituting this into the second equation: $5 = 9 - 2b$, from which we find $b = 2$. Finally, we get $a = 1$. The desired remainder is $r(x) = x + 2$.

![](https://cdn.mathpix.com/cropped/2024_06_07_fb5898f0472fc84ca82bg-11.jpg?height=60&width=1633&top_left_y=1004&top_left_x=220)

Conclusion: the degree of the remainder $r(x)$ is $\mathbf{1}$ or less, $r(x) = ax + b \ldots \ldots . .2$ points

Using the fact that $p(\alpha)$ is the remainder when dividing the polynomial $p(x) \mathbf{by} x - \alpha \ldots \ldots \mathbf{1}$ point

Setting up a system of two equations with two unknowns and solving it ..... 3 points