14.2k views
5 votes
Let f(x) = n^2 - 1, where n is an integer and n >= 1. Is f(x) always divisible

by (x - 1)? Justify.

User Tom Lord
by
8.2k points

1 Answer

5 votes

Answer:

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.

User Kartick Shaw
by
6.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories