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Difference between revisions of "1965 AHSME Problems/Problem 27"

(Created page with "Let <math>h(y)=y^2+my+2</math> <math>h(y)=y^2+my+2=f(y)(y-1)R_1</math> h(y)=<math>y^2</math>+my+2=g(y)(y+1)<math>R_2</math> h(1)=3+m=<math>R_1</math> h(-1)=3-m=<math>R_2</...")
 
(Add statement & Unify answer)
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== Problem ==
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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>.
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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:
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<math>\textbf{(A)}\ 0 \qquad
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\textbf{(B) }\ 1 \qquad
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\textbf{(C) }\ 2 \qquad
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\textbf{(D) }\ - 1 \qquad
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\textbf{(E) }\ \text{an undetermined constant} </math>
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== Solution ==
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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
  
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The answer is A
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The answer is <math>\boxed{A}</math>

Revision as of 13:53, 16 July 2024

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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