Chapter 2.3, Problem 115E

To determine

To complete: Each polynomial division, to generalize the pattern, to find a formula for the polynomial division $\frac{x^n-1}{x-1}$ and to create a numerical example to test this formula.

Answer to Problem 115E

  $\begin{array}{l}(a)\frac{x^2-1}{x-1}=x+1(x\ne 1)\\ (b)\frac{x^3-1}{x-1}=x^2+x+1(x\ne 1)\\ (c)\frac{x^4-1}{x-1}=x^3+x^2+x+1(x\ne 1)\\ \\ Generalformulais:\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\mathrm{...}+x+1(x\ne 1).\end{array}$

Explanation of Solution

Given information: The following divisions are to be done:

  $\begin{array}{l}(a)\frac{x^2-1}{x-1}\\ (b)\frac{x^3-1}{x-1}\\ (c)\frac{x^4-1}{x-1}\end{array}$

Calculation:

Let us perform the divisions by synthetic division.

  $(a)\frac{x^2-1}{x-1}$

Here, $x^2-1$ is the dividend and $x-1$ is the divisor.

  $\begin{array}{llll}1\hfill & 1\hfill & 0\hfill & -1\hfill \\ \hfill & \hfill & 1\hfill & 1\hfill \\ \hfill & 1\hfill & 1\hfill & 0\hfill \end{array}$

So, the quotient is $(x+1)$ and the remainder is $0$ .

Thus, $\frac{x^2-1}{x-1}=x+1(x\ne 1)$ .

  $(b)\frac{x^3-1}{x-1}$

Here, $x^3-1$ is the dividend and $x-1$ is the divisor.

  $\begin{array}{lllll}1\hfill & 1\hfill & 0\hfill & 0\hfill & -1\hfill \\ \hfill & \hfill & 1\hfill & 1\hfill & 1\hfill \\ \hfill & 1\hfill & 1\hfill & 1\hfill & 0\hfill \end{array}$

So, the quotient is $(x^2+x+1)$ and the remainder is $0$ .

Thus, $\frac{x^3-1}{x-1}=x^2+x+1(x\ne 1)$ .

  $(c)\frac{x^4-1}{x-1}$

Here, $x^4-1$ is the dividend and $x-1$ is the divisor.

  $\begin{array}{llllll}1\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill & -1\hfill \\ \hfill & \hfill & 1\hfill & 1\hfill & 1\hfill & 1\hfill \\ \hfill & 1\hfill & 1\hfill & 1\hfill & 1\hfill & 0\hfill \end{array}$

So, the quotient is $(x^3+x^2+x+1)$ and the remainder is $0$ .

Thus, $\frac{x^4-1}{x-1}=x^3+x^2+x+1(x\ne 1)$ .

Generalizing the pattern of the quotient,

  $\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\mathrm{...}+x+1(x\ne 1)$ .

Hence, the general formula for the polynomial division $\frac{x^n-1}{x-1}$ is $x^{n-1}+x^{n-2}+x^{n-3}+\mathrm{...}+x+1$ .

Thus, $\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\mathrm{...}+x+1(x\ne 1)$ .

Putting $n=2$ in this formula, $\frac{x^2-1}{x-1}=x+1$ , which matches with the result in part $(a)$ .

Hence, the formula is verified.