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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</...")
 
(Added solution to the problem)
 
(3 intermediate revisions by 2 users not shown)
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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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<math>h(y)=y^2+my+2=f(y)(y-1)R_1</math>
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<math>h(y)=y^2+my+2=f(y)(y-1)+R_1</math>
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<math>h(y)=y^2+my+2=g(y)(y+1)+R_2</math>
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<math>h(1)=3+m=R_1</math>
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<math>h(-1)=3-m=R_2</math>
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<math>m=0</math>
  
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h(y)=<math>y^2</math>+my+2=g(y)(y+1)<math>R_2</math>
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Thus, the answer is <math>\boxed{\textbf{(A) }0}</math>.
  
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h(1)=3+m=<math>R_1</math>
 
  
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h(-1)=3-m=<math>R_2</math>
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== Solution 2 ==
  
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m=0
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As in the other solution, let <math>h(y)=y^2+my+2</math>. By dividing <math>h(y)</math> by <math>(y-1)</math> (through either long or [[synthetic division|synthetic]] division), we get a remainder <math>R_1=m+3</math>. Similarly, dividing <math>h(y)</math> by <math>(y+1)</math> yields the remainder <math>R_2=3-m</math>. Setting <math>R_1=R_2</math>, we see that <math>m+3=3-m</math>, which only holds when <math>m=0</math>, which corresponds to answer choice <math>\fbox{\textbf{(A)}}</math>.
  
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The answer is A
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== See Also ==
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{{AHSME 40p box|year=1965|num-b=26|num-a=28}}
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{{MAA Notice}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 19:21, 18 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 $h(y)=y^2+my+2$

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

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

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

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

$m=0$

Thus, the answer is $\boxed{\textbf{(A) }0}$.


Solution 2

As in the other solution, let $h(y)=y^2+my+2$. By dividing $h(y)$ by $(y-1)$ (through either long or synthetic division), we get a remainder $R_1=m+3$. Similarly, dividing $h(y)$ by $(y+1)$ yields the remainder $R_2=3-m$. Setting $R_1=R_2$, we see that $m+3=3-m$, which only holds when $m=0$, which corresponds to answer choice $\fbox{\textbf{(A)}}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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