The sum of all possible values of n, where n is an integer when 4n^2 + 6n + 8 is divided by 2n gives an integer, is
Answers
Given : 4n² + 6n + 8 is divided by 2n gives an integer
To find : sum of all possible values of n
Solution:
4n² + 6n + 8 is divisible by 2n
2n _| 4n² + 6n + 8 |_ 2n + 3
4n²
_____
6n + 8
6n
________
8
8 must be divisible by 2n
8 is divisible by 1 , 2 , 4 or 8
2n = 1 does not give integral value of n
2n = 2 , 4 , 8
=> n = 1 , 2 , 4
Possible values of n = 1 , 2 , 4
Sum = 1 + 2 + 4 = 7
or simply 4n² + 6n + 8 = 2n(2n + 3) + 8 hence 8 must be divisible by 2n
sum of all possible values of n = 7
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