Chapter 2.4, Problem 12E

Solution

To divide the polynomial $x^8-1$ with $x+2$ and to summarize the statement in fraction form.

It is found that $x^8-1$ gives the quotient $x^7-2x^6+4x^5-8x^4+16x^3-32x^2+64x-128$ and the remainder $255$ when it is divided by $x+2$ .

Thus, it is found that:

  $\frac{x^8-1}{x+2}=\left(x^7-2x^6+4x^5-8x^4+16x^3-32x^2+64x-128\right)+\frac{255}{x+2}$ .

Given:

The dividend:

  $x^8-1$

The divisor:

  $x+2$

Set up:

The zero of the divisor $x+2$ is $-2$ . So, put $-2$ in the divisor position.

The dividend $x^8-1$ is not in the standard form.

Thus, write it in the standard form at first:

  $x^8+0x^7+0x^6+0x^5+0x^4+0x^3+0x^2+0x-1$

Then, write its coefficients in the order in the dividend position.

  $\begin{array}{cc}\overline{)-2}& \underset{¯}{\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& -1\\ & & & & & & & & \end{array}}\\ & \begin{array}{ccccccccc}& & & & & & & & \end{array}\end{array}$ .

Calculation:

Because the leading coefficient of the dividend must be the leading coefficient of the quotient, write $1$ in the first quotient position.

  $\begin{array}{cc}\overline{)-2}& \underset{¯}{\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& -1\\ & & & & & & & & \end{array}}\\ & \begin{array}{ccccccccc}1& & & & & & & & \end{array}\end{array}$ .

Multiply the zero of the divisor, $-2$ , by the most recently determined coefficient of the quotient , $1$ . Now, write this product above the line and one column to the right.

  $\begin{array}{cc}\overline{)-2}& \underset{¯}{\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& -1\\ & -2& & & & & & & \end{array}}\\ & \begin{array}{ccccccccc}1& & & & & & & & \end{array}\end{array}$ .

Add the next coefficient of the dividend with the product and record the sum below the line.

  $\begin{array}{cc}\overline{)-2}& \underset{¯}{\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& -1\\ & -2& & & & & & & \end{array}}\\ & \begin{array}{ccccccccc}1& -2& & & & & & & \end{array}\end{array}$ .

Repeat the process until the end.

  $\begin{array}{cc}\overline{)-2}& \underset{¯}{\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& -1\\ & -2& 4& -8& 16& -32& 64& -128& 256\end{array}}\\ & \begin{array}{ccccccccc}1& -2& 4& -8& 16& -32& 64& -128& 255\end{array}\end{array}$ .

Interpret:

The numbers below the line give the coefficients of the quotient and the remainder.

The first eight numbers ( $1$ , $-2$ , $4$ , $-8$ , $16$ , $-32$ , $64$ and $-128$ ) form the coefficients of the quotient, a seven degree polynomial: $\left(1\right)x^7+\left(-2\right)x^6+\left(4\right)x^5+\left(-8\right)x^4+\left(16\right)x^3+\left(-32\right)x^2+\left(64\right)x+\left(-128\right)$ .

The last number $255$ is the remainder.

Thus, the quotient is $x^7-2x^6+4x^5-8x^4+16x^3-32x^2+64x-128$ and the remainder is $255$ .

Conclusion:

It is found that $x^8-1$ gives the quotient $x^7-2x^6+4x^5-8x^4+16x^3-32x^2+64x-128$ and the remainder $255$ when it is divided by $x+2$ .

Thus, it is found that:

  $\frac{x^8-1}{x+2}=\left(x^7-2x^6+4x^5-8x^4+16x^3-32x^2+64x-128\right)+\frac{255}{x+2}$ .