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If f (x)=x^99+ 99x + k is divisible by x + 1 , then k = A. −100 B. −98 C. 98...
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`f(x) = x^99 +99x+k`
Using remainder theorem if fully `f(x)` is fully divided by ` (x-a)` then` f(a) = 0` ;
`f(-1) = 0`
`f(-1) = -(1)^99+99(-1)+k = 0`
`-1-99+k = 0`
`K = 100`
So the answer is k = 100. Correct answer is at option D).
Posted by jeew-m on September 4, 2013 at 6:04 PM (Answer #1)
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