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

(Fixed page formatting)
Line 13: Line 13:
 
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>.  
 
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>.  
 
Thus the answer is <math>2x -2</math>, for choice <math>\boxed{(E)}</math>.
 
Thus the answer is <math>2x -2</math>, for choice <math>\boxed{(E)}</math>.
 +
 +
==Solution 2==
 +
 +
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>
  
 
==See Also==
 
==See Also==

Revision as of 11:41, 11 September 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 $E$

See Also

1982 AHSME (ProblemsAnswer KeyResources)
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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png