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

 May 25, 2022
 #1
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f(x)=x^(101) + Ax + B

 

x^2-1=(x-1)(x+1)  so  1 and -1 are roots of f(x)

 

so

f(1)=0    and  f(-1)=0

 

solve simultaneously.

 

Easy peazy.

 May 25, 2022

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