Question

(x+1) is a factor of xⁿ+1 only if
A. n is an odd integer
B. n is an even integer
C. n is a negative integer
D. n is a positive integer

Answer 1

Given: (x + 1) is a factor of xⁿ + 1  

 

Solution :  

Let f (x) = xⁿ +1

Since, x + 1 is a factor of f (x), then f (- 1) = 0

On Substituting x = - 1 in f (x),

f (-1) = xⁿ +1

f (-1) = (-1)ⁿ + 1

f (-1) = (-1)¹ + 1

f (-1) = - 1 + 1

f (-1) = 0

Here 1 is an odd integer  

If we take n as an even integer then f (-1) ≠ 0.

Hence, n is an odd integer.

Among the given options option (A) n is an odd integer is correct.

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Answer 2

Given that:

x + 1 is a factor of f(x) = xⁿ + 1

To find:

Value of n

Solution:

If x + 1 is a factor then:

x + 1 = 0

x = -1

When we put this value of x in f(x) = xⁿ + 1, we should get the result equal to 0, since it's given that x + 1 is a factor of f(x).

f(-1) = 0

(-1)ⁿ + 1 = 0

(-1)ⁿ = -1

When bases are same, the powers are kept equal. Thus:

n = 1

We know that, 1 is an odd integer.

Thus, option (A) is correct.