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Difference between revisions of "1982 AHSME Problems/Problem 1"

(Created page with "By simply dividing <math>\frac{x^3-2}{x^2-2}</math> by using polynomial long division, you get : <math>x + \frac{2x-2}{x^2-2}</math>. Thus our answer will be <math>2x -2</mat...")
 
(→‎Solution 2)
 
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By simply dividing <math>\frac{x^3-2}{x^2-2}</math> by using polynomial long division, you get : <math>x + \frac{2x-2}{x^2-2}</math>.  
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==Problem==
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Thus our answer will be <math>2x -2</math>, for choice <math>\fbox{E}</math>
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When the polynomial <math>x^3-2</math> is divided by the polynomial <math>x^2-2</math>, the remainder is
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<math>\text{(A)} \ 2 \qquad
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\text{(B)} \ -2 \qquad
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\text{(C)} \ -2x-2 \qquad
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\text{(D)} \ 2x+2 \qquad
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\text{(E)} \ 2x-2</math> 
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==Solution==
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Working out <math>\frac{x^3-2}{x^2-2}</math> using polynomial long division, we get <math>x + \frac{2x-2}{x^2-2}</math>.  
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Thus the answer is <math>2x -2</math>, for choice <math>\boxed{(E)}</math>.
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==Solution 2==
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Substituting an easy value, like <math>3</math>, you get <math>\frac{25}{7}</math>, with a remainder of 4. Then you just plug in x to get <math>E</math>
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-dragoon
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==See Also==
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{{AHSME box|year=1982|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 11:27, 10 October 2022

Problem

When the polynomial $x^3-2$ is divided by the polynomial $x^2-2$, the remainder is

$\text{(A)} \ 2 \qquad \text{(B)} \ -2 \qquad \text{(C)} \ -2x-2 \qquad \text{(D)} \ 2x+2 \qquad \text{(E)} \ 2x-2$

Solution

Working out $\frac{x^3-2}{x^2-2}$ using polynomial long division, we get $x + \frac{2x-2}{x^2-2}$. Thus the answer is $2x -2$, for choice $\boxed{(E)}$.

Solution 2

Substituting an easy value, like $3$, you get $\frac{25}{7}$, with a remainder of 4. Then you just plug in x to get $E$

-dragoon

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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
All AHSME Problems and Solutions


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