202111221509 Solution to 1971-CE-AMATH-1

When the functions $x^2-bx+3$ and $2b-x$ are divided by $(x-a)$ , the remainders are 1 and 4 respectively. Find $a$ and $b$ .

Solution.

Translating mathematically, we have

$\begin{aligned} x^2-bx+3 & = (x-a)F(x)+1 \\ 2b-x & = (x-a)G(x)+4\\ \end{aligned}$

Plugging in $x=a$ , we have

$\begin{aligned} (a)^2-b(a)+3 & = \big( (a)-a\big) F(a)+1 \\ 2b-(a) & = \big( (a)-a\big) G(a)+4 \\ \end{aligned}$

After simplifying it, we have

$\begin{aligned} a^2-ab & = -2\\ -a+2b & = 4\\ \end{aligned}$

Solving two equations with two unknowns,

$\begin{cases} & a =2\\ & b =3 \\ \end{cases}$

is the solution.