Difference between revisions of "1987 AHSME Problems/Problem 1"

(Created page with "==Problem== <math>(1+x^2)(1-x^3)</math> equals <math>\text{(A)}\ 1 - x^5\qquad \text{(B)}\ 1 - x^6\qquad \text{(C)}\ 1+ x^2 -x^3\qquad \\ \text{(D)}\ 1+x^2-x^3-x^5\qquad \tex...")
 
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\text{(C)}\ 1+ x^2 -x^3\qquad \\  
 
\text{(C)}\ 1+ x^2 -x^3\qquad \\  
 
\text{(D)}\ 1+x^2-x^3-x^5\qquad
 
\text{(D)}\ 1+x^2-x^3-x^5\qquad
\text{(E)}\ 1+x^2-x^3-x^6 </math>  
+
\text{(E)}\ 1+x^2-x^3-x^6 </math>
  
 +
==Solution==
  
 +
We multiply: <math>(1+x^2)(1-x^3) = 1 - x^3 + x^2 - x^5</math>. Thus the answer is <math>\boxed{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:33, 23 April 2017

Problem

$(1+x^2)(1-x^3)$ equals

$\text{(A)}\ 1 - x^5\qquad \text{(B)}\ 1 - x^6\qquad \text{(C)}\ 1+ x^2 -x^3\qquad \\  \text{(D)}\ 1+x^2-x^3-x^5\qquad \text{(E)}\ 1+x^2-x^3-x^6$

Solution

We multiply: $(1+x^2)(1-x^3) = 1 - x^3 + x^2 - x^5$. Thus the answer is $\boxed{D}$

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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