$29 \cdot 42 y^{2}+m y+2$ divided by $y-1$ gives the quotient $f(y)$, the remainder is $R_{1}, y^{2}+m y+$ 2 divided by $y+1$ gives the quotient $g(y)$, the remainder is $R_{2}$, if $R_{1}=R_{2}$, then $m$ equals
(A) 0 .
(B) 1 .
(C) 2 .
(D) -1 .
(E) an indeterminate number.
(16th American High School Mathematics Examination, 1965)
[Solution] Given $y^{2}+m y+2=(y-1) f(y)+R_{1}$,
and $y^{2}+m y+2=(y+1) g(y)+R_{2}$.
In (1), let $y=1$, we get $R_{1}=3+m$.
In (2), let $y=-1$, we get $R_{2}=3-m$.
Given $R_{1}=R_{2}$, hence $3+m=3-m$, we get $m=0$. Therefore, the answer is $(A)$.