1965 AHSME Problems/Problem 27

Problem

When $y^2 + my + 2$ is divided by $y - 1$ the quotient is $f(y)$ and the remainder is $R_1$ . When $y^2 + my + 2$ is divided by $y + 1$ the quotient is $g(y)$ and the remainder is $R_2$ . If $R_1 = R_2$ then $m$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ - 1 \qquad \textbf{(E) }\ \text{an undetermined constant}$

Solution

Let $h(y)=y^2+my+2$

$h(y)=y^2+my+2=f(y)(y-1)R_1$

h(y)= $y^2$ +my+2=g(y)(y+1) $R_2$

h(1)=3+m= $R_1$

h(-1)=3-m= $R_2$

m=0

The answer is $\boxed{A}$

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1965 AHSME Problems/Problem 27

Problem

Solution