202212211204 Solution to 1972-CE-AMATH-I-X

$F(x)\equiv A(x-1)^2+B(x-1)+C$ where $A$ , $B$ , and $C$ are constants. Find the values of $A$ , $B$ , and $C$ , given that $(x+2)$ is a factor of $F(x)$ and that the remainders when $F(x)$ is divided by $(x-1)$ and $(x+1)$ are $9$ and $-11$ respectively.

Roughwork.

$\begin{aligned} & \quad\enspace (x+2)\textrm{ is a factor of }F(x) \\ & \Leftrightarrow F(x)=(x+2)\cdot G(x) \\ & \Leftrightarrow F(-2)=0 \\ & \Leftrightarrow A((-2)-1)^2+B((-2)-1)+C=0 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{9}\textrm{ remains when dividing }F(x)\textrm{ by }(x-1) \\ &\Leftrightarrow F(1)=C=9 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{-11}\textrm{ remains when dividing }F(x)\textrm{ by }(x+1) \\ &\Leftrightarrow F(-1)=A((-1)-1)^2+B((-1)-1)+C=-11 \\ \end{aligned}$

This problem is not to be attempted.

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