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The polynomial \(x^{101} + Ax + B\) is divisible by \(x^2 + x + 1\)  for some real numbers \(A\) and \(B\). Find \(A+B\)

 Mar 20, 2020
 #1
avatar+1116 
+2

https://web2.0calc.com/questions/sorry-for-re-posting-but-i-really-need-an-answer

 Mar 20, 2020
 #2
avatar+24925 
+2

The polynomial x^{101} + Ax + B  is divisible by x^2 + x + 1 for some real numbers A and B.
Find A+B

 

answer see: https://web2.0calc.com/questions/sorry-for-re-posting-but-i-really-need-an-answer#r2

 

laugh

 Mar 20, 2020

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