Chapter 1.3, Problem 136E

To determine

To verify: The results provided below,

  $\begin{array}{l}{A}^2-1=\left(A-1\right)\left(A+1\right)\\ A^3-1=\left(A-1\right)\left(A^2+A+1\right)\\ A^4-1=\left(A-1\right)\left(A^3+A^2+A+1\right)\end{array}$

Also explain the factorization of $A^5-1$ . On a similar pattern obtain the factorization of $A^n-1$ , where n is a positive integer.

Explanation of Solution

Given information:

The expressions,

  $\begin{array}{l}{A}^2-1=\left(A-1\right)\left(A+1\right)\\ A^3-1=\left(A-1\right)\left(A^2+A+1\right)\\ A^4-1=\left(A-1\right)\left(A^3+A^2+A+1\right)\end{array}$

Proof:

Consider the expressions,

  $\begin{array}{l}{A}^2-1=\left(A-1\right)\left(A+1\right)\\ A^3-1=\left(A-1\right)\left(A^2+A+1\right)\\ A^4-1=\left(A-1\right)\left(A^3+A^2+A+1\right)\end{array}$

Now, take one expression at a time and expand their right hand side.

For $A^2-1=\left(A-1\right)\left(A+1\right)$ ,

  $\begin{array}{c}\left(A-1\right)\left(A+1\right)=A^2+A-A-1\\ =A^2-1\end{array}$

For $A^3-1=\left(A-1\right)\left(A^2+A+1\right)$ ,

  $\begin{array}{c}\left(A-1\right)\left(A^2+A+1\right)=A^3+A^2+A-A^2-A-1\\ =A^3-1\end{array}$

For $A^4-1=\left(A-1\right)\left(A^3+A^2+A+1\right)$ ,

  $\begin{array}{c}\left(A-1\right)\left(A^3+A^2+A+1\right)=A^4+A^3+A^2+A-A^3-A^2-A-1\\ =A^4-1\end{array}$

Observe the pattern to factorize the polynomial $A^5-1$ , first factor is of the form $\left(A-1\right)$ and second factor is a polynomial with degree one less than the polynomial.

  $A^5-1=\left(A-1\right)\left(A^4+A^3+A^2+A+1\right)$

Now, verify the expression obtained above,

  $\begin{array}{c}\left(A-1\right)\left(A^4+A^3+A^2+A+1\right)=A^5+A^4+A^3+A^2+A-A^3-A^2-A-1\\ =A^5-1\end{array}$

Observe the pattern to factorize the polynomial $A^n-1$ , first factor is of the form $\left(A-1\right)$ and second factor is a polynomial with degree one less than the polynomial.

  $A^n-1=\left(A-1\right)\left(A^{n-1}+A^{n-1}+\cdots +A^2+A+1\right)$

Hence, the factorization of the polynomials of the form $A^n-1$ has been verified.