Chapter 2.4, Problem 7E

Solution

To divide the polynomial $x^3-5x^2+3x-2$ with $x+1$ and to summarize the statement in fraction form.

It is found that $x^3-5x^2+3x-2$ gives the quotient $x^2-6x+9$ and the remainder $-11$ when it is divided by $x+1$ .

Thus, it is found that:

  $\frac{x^3-5x^2+3x-2}{x+1}=\left(x^2-6x+9\right)-\frac{11}{x+1}$ .

Given:

The dividend:

  $x^3-5x^2+3x-2$

The divisor:

  $x+1$

Set up:

The zero of the divisor $x+1$ is $-1$ . So, put $-1$ in the divisor position.

The dividend $x^3-5x^2+3x-2$ is in the standard form. Thus, write its coefficients in the order in the dividend position.

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 1& \hfill -5& \hfill 3& \hfill -2\\ \hfill & \hfill & \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill & \hfill & \hfill & \hfill \end{array}\end{array}$

Calculation:

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

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 1& \hfill -5& \hfill 3& \hfill -2\\ \hfill & \hfill & \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 1& \hfill & \hfill & \hfill \end{array}\end{array}$ .

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

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 1& \hfill -5& \hfill 3& \hfill -2\\ \hfill & \hfill -1& \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 1& \hfill & \hfill & \hfill \end{array}\end{array}$ .

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

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 1& \hfill -5& \hfill 3& \hfill -2\\ \hfill & \hfill -1& \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 1& \hfill -6& \hfill & \hfill \end{array}\end{array}$ .

Repeat the process until the end.

  $\begin{array}{cc}\overline{)-1}& \underset{¯}{\begin{array}{rrrr}\hfill 1& \hfill -5& \hfill 3& \hfill -2\\ \hfill & \hfill -1& \hfill 6& \hfill -9\end{array}}\\ & \begin{array}{rrrr}\hfill 1& \hfill -6& \hfill 9& \hfill -11\end{array}\end{array}$ .

Interpret:

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

The first three numbers ( $1$ , $-6$ and $9$ ) form the coefficients of the quotient, a second degree polynomial: $\left(1\right)x^2+\left(-6\right)x+\left(9\right)$ .

The last number $-11$ is the remainder.

Thus, the quotient is $x^2-6x+9$ and the remainder is $-11$ .

Conclusion:

It is found that $x^3-5x^2+3x-2$ gives the quotient $x^2-6x+9$ and the remainder $-11$ when it is divided by $x+1$ .

Thus, it is found that:

  $\frac{x^3-5x^2+3x-2}{x+1}=\left(x^2-6x+9\right)-\frac{11}{x+1}$ .