1965 AHSME Problems/Problem 27: Difference between revisions
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== Problem == | |||
When <math>y^2 + my + 2</math> is divided by <math>y - 1</math> the quotient is <math>f(y)</math> and the remainder is <math>R_1</math>. | |||
When <math>y^2 + my + 2</math> is divided by <math>y + 1</math> the quotient is <math>g(y)</math> and the remainder is <math>R_2</math>. If <math>R_1 = R_2</math> then <math>m</math> is: | |||
<math>\textbf{(A)}\ 0 \qquad | |||
\textbf{(B) }\ 1 \qquad | |||
\textbf{(C) }\ 2 \qquad | |||
\textbf{(D) }\ - 1 \qquad | |||
\textbf{(E) }\ \text{an undetermined constant} </math> | |||
== Solution == | |||
Let <math>h(y)=y^2+my+2</math> | Let <math>h(y)=y^2+my+2</math> | ||
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m=0 | m=0 | ||
The answer is A | The answer is <math>\boxed{A}</math> | ||
Revision as of 12:53, 16 July 2024
Problem
When 
is divided by 
the quotient is 
and the remainder is 
.
When is divided by 
the quotient is 
and the remainder is 
. If 
then 
is:
Solution
Let 
h(y)=
+my+2=g(y)(y+1)
h(1)=3+m=
h(-1)=3-m=
m=0
The answer is 
