Question

Which of the following is a factor of \(n(n^2-1)\) for all \(n > 1\)?

A. n+1
B. n-1
C. n+2
D. n-2

Answer 1

For the expression n(n^2 - 1), the term (n^2 - 1) is a difference of squares and can be factored into (n + 1)(n - 1), making n - 1 a factor for all n > 1.

Step-by-step explanation:

To determine which of the options is a factor of the expression n(n^2 - 1) for all n > 1, we need to recognize that n^2 - 1 is a difference of squares. It can be factored into (n + 1)(n - 1).

Therefore, the expression n(n^2 - 1) can be rewritten as n(n + 1)(n - 1). This shows that both n + 1 and n - 1 are factors of the given expression for any n > 1. Looking at the options provided, B. n - 1 is always a factor of the expression.

Answer 2

The correct factor of n(n²-1) for all n > 1 is B. n-1.

Step-by-step explanation:

To understand why n-1 is a factor of n(n²-1) for all n > 1, let's factorize n(n²-1). First, notice that n(n²-1) can be expressed as n(n+1)(n-1). This factorization is derived from the difference of squares identity, where n²-1 = (n+1)(n-1). Now, we can see that n-1 is a factor of n(n²-1) since it appears in the factorization.

Let's consider a quick example to illustrate this. If we take n = 3, then n(n+1)(n-1) = 3(4)(2) = 24. In this case, n-1 is 2, and it is indeed a factor of 24. This example demonstrates that for any n > 1, n-1 will be a factor of n(n²-1).

In conclusion, by factoring n(n²-1) and understanding the properties of the difference of squares, we can clearly see that the correct factor among the given options is n-1.