Chapter 8.4, Problem 10WE

To determine

To divide : the given polynomial by the binomial using synthetic division.

Answer to Problem 10WE

  $\frac{x^6-1}{x+1}=x^5-x^4+x^3-x^2+x-1$

Explanation of Solution

Given information :

The polynomial: $x^6-1$

The binomial: $x+1$

Formula used : The divisor in a synthetic division is a binomial of the form $x-c$ .

Calculation :

As per the problem,

The dividend is $x^6-1$ . Insert the missing terms, $0x^5,0x^4,0x^3,0x^2\text{, and 0}x$ .

Since the divisor is $x+1=x-(-1)$ , use $-1$ for c .

  $\begin{array}{l}\overline{)-1}\text{ 1 0 0 }00\text{ 0 }-\text{1}\\ \underset{\_}{-\text{1 1 }-\text{1 1 }-\text{1 1}}\\ \text{ 1 }-1\text{ 1 }-\text{1 1 }-1\text{ | 0}\end{array}$

The quotient is $x^5-x^4+x^3-x^2+x-1$ , and the remainder is $0$

Therefore,

  $\frac{x^6-1}{x+1}=x^5-x^4+x^3-x^2+x-1$