2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(4)
Problem 1(4)
The division of a polynomial function $f(x)$ by $(x-1)^2$ gives the remainder of $x+1$, and that by $x^2$ gives the remainder $2x+3$. Thus, the remainder of the division of $f(x)$ by $x^2(x-1)$ is $$ \fbox{ A }x^2 + \fbox{ B }x + \fbox{ C } . $$
Solution
We need to find the remainder when $f(x)$ is divided by $x^2(x-1)$. Because $x^2(x-1)$ is of order $n=3$, the remainder will be of at most the order $n=2$, which means that it is of the form $Ax^2 + Bx + C$. The problem is now to find the coefficients $A,B$ and $C$ such that $$ f(x) = Q(x)\cdot x^2(x-1) + (Ax^2 + Bx + C) $$ where $Q(x)$ is the quotient and $(Ax^2 + Bx + C)$ is the remainder.
We first write down the given conditions.
- Division of $f(x)$ by $(x-1)^2$:
$$
f(x) = q_1(x)\cdot(x-1)^2 + (x+1),
$$
where $q_1(x)$ is the quotient and $x+1$ is the remainder.
From this, we know that $$ (★):f(1) = q_2(x)\cancelto{0}{(1-1)} + (1+1) = 2. $$
- Division of $f(x)$ by $x^2$:
$$
(★★):f(x) = q_2(x)\cdot x^2 + (2x+3),
$$
where $q_2(x)$ is the quotient and $2x+3$ is the remainder.
We can also write this using $Q(x)$ and grouping $Ax^2$ with it so that we get $$ (★★★):f(x) = \left(Q(x)\cdot (x-1) + A\right)x^2 + (Bx + C), $$ which we can compare with (★★)to conclude that for all values of $x$, $$ 2x + 3 = Bx + C \Leftrightarrow B = 2, C =3. $$ Furthermore, we can set $x=1$ in (★★★) and use the values of $B = 2$ and $C =3$ to obtain $$\begin{align} f(1) = \left(Q(x)\cdot \cancelto{0}{(1-1)} + A\right) \cdot 1^2 + (2\cdot 1 + 3) = A + 5, \end{align}$$ which we can now compare with (★)to obtain $$ \begin{align} 2 = f(1) &= A + 5\\ A &= -3. \end{align}$$
Therefore, the coefficients are $A=-3, B=2$ and $C=3$.■
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