Question

Let f(x) = n^2 - 1, where n is an integer and n >= 1. Is f(x) always divisible

by (x - 1)? Justify.

Answer 1

Yes.

Explanation:

(Assume x is not 1.)

`(x^n-1)/(x-1)` is always divisible for integers greater than 1.

Let's use synthetic division:

1 | 1x^n + 0x^(n-1) +0x^(n-2) + ....+0x^3+0x^2+0x -1

| 1 1 1 1 1 1

---------------------------------------------------------------------------------

1 1 1 1 1 1 0

We see the remainder is 0 which means that
`(x-1)` divides
`(x^n-1)`.

The quotient is
`x^(n-1)+x^(n-2)+x^(n-3)+\cdots+x^2+x+1`.