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

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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==
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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==See Also==
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{{AHSME box|year=1982|before=First Question|num-a=2}}
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{{MAA Notice}}

Revision as of 23:01, 20 February 2019

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 $\boxed{(E)}$.

See Also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AHSME Problems and Solutions


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