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)
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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