Chapter 2.4, Problem 9E

Solution

To divide the polynomial $9x^3+7x^2-3x$ with $x-10$ and to summarize the statement in fraction form.

It is found that $9x^3+7x^2-3x$ gives the quotient $9x^2+97x+967$ and the remainder $9670$ when it is divided by $x-10$ .

Thus, it is found that:

  $\frac{9x^3+7x^2-3x}{x-10}=\left(9x^2+97x+967\right)+\frac{9670}{x-10}$ .

Given:

The dividend:

  $9x^3+7x^2-3x$

The divisor:

  $x-10$

Set up:

The zero of the divisor $x-10$ is $10$ . So, put $10$ in the divisor position.

The dividend $9x^3+7x^2-3x$ is not in the standard form.

Thus, write it in the standard form at first:

  $9x^3+7x^2-3x+0$

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

  $\begin{array}{cc}\overline{)10}& \underset{¯}{\begin{array}{rrrr}\hfill 9& \hfill 7& \hfill -3& \hfill 0\\ \hfill & \hfill & \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill & \hfill & \hfill & \hfill \end{array}\end{array}$ .

Calculation:

Because the leading coefficient of the dividend must be the leading coefficient of the quotient, write $9$ in the first quotient position.

  $\begin{array}{cc}\overline{)10}& \underset{¯}{\begin{array}{rrrr}\hfill 9& \hfill 7& \hfill -3& \hfill 0\\ \hfill & \hfill & \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 9& \hfill & \hfill & \hfill \end{array}\end{array}$ .

Multiply the zero of the divisor, $10$ , by the most recently determined coefficient of the quotient , $9$ . Now, write this product above the line and one column to the right.

  $\begin{array}{cc}\overline{)10}& \underset{¯}{\begin{array}{rrrr}\hfill 9& \hfill 7& \hfill -3& \hfill 0\\ \hfill & \hfill 90& \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 9& \hfill & \hfill & \hfill \end{array}\end{array}$ .

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

  $\begin{array}{cc}\overline{)10}& \underset{¯}{\begin{array}{rrrr}\hfill 9& \hfill 7& \hfill -3& \hfill 0\\ \hfill & \hfill 90& \hfill & \hfill \end{array}}\\ & \begin{array}{rrrr}\hfill 9& \hfill 97& \hfill & \hfill \end{array}\end{array}$ .

Repeat the process until the end.

  $\begin{array}{cc}\overline{)10}& \underset{¯}{\begin{array}{rrrr}\hfill 9& \hfill 7& \hfill -3& \hfill 0\\ \hfill & \hfill 90& \hfill 970& \hfill 9670\end{array}}\\ & \begin{array}{rrrr}\hfill 9& \hfill 97& \hfill 967& \hfill 9670\end{array}\end{array}$ .

Interpret:

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

The first three numbers ( $9$ , $97$ and $967$ ) form the coefficients of the quotient, a second degree polynomial: $\left(9\right)x^2+\left(97\right)x+\left(967\right)$ .

The last number $9670$ is the remainder.

Thus, the quotient is $9x^2+97x+967$ and the remainder is $967$ .

Conclusion:

It is found that $9x^3+7x^2-3x$ gives the quotient $9x^2+97x+967$ and the remainder $9670$ when it is divided by $x-10$ .

Thus, it is found that:

  $\frac{9x^3+7x^2-3x}{x-10}=\left(9x^2+97x+967\right)+\frac{9670}{x-10}$ .