Chapter 2.4, Problem 8E

Solution

To divide the polynomial $2x^4-5x^3+7x^2-3x+1$ by $x-3$ and to summarize the statement in fraction form.

It is found that $2x^4-5x^3+7x^2-3x+1$ gives the quotient $2x^3+x^2+10x+27$ and the remainder $82$ when it is divided by $x-3$ .

Thus, it is found that:

  $\frac{2x^4-5x^3+7x^2-3x+1}{x-3}=\left(2x^3+x^2+10x+27\right)+\frac{82}{x-3}$ .

Given:

The dividend:

  $2x^4-5x^3+7x^2-3x+1$

The divisor:

  $x-3$

Set up:

The zero of the divisor $x-3$ is $3$ . So, put $3$ in the divisor position.

The dividend $2x^4-5x^3+7x^2-3x+1$ is in the standard form. Thus, write its coefficients in the order in the dividend position.

  $\begin{array}{cc}\overline{)3}& \underset{¯}{\begin{array}{ccccc}2& -5& 7& -3& 1\\ & & & & \end{array}}\\ & \begin{array}{ccccc}& & & & \end{array}\end{array}$ .

Calculation:

Because the leading coefficient of the dividend must be the leading coefficient of the quotient, write $2$ in the first quotient position.

  $\begin{array}{cc}\overline{)3}& \underset{¯}{\begin{array}{ccccc}2& -5& 7& -3& 1\\ & & & & \end{array}}\\ & \begin{array}{ccccc}2& & & & \end{array}\end{array}$ .

Multiply the zero of the divisor, $3$ , by the most recently determined coefficient of the quotient , $2$ . Now, write this product above the line and one column to the right.

  $\begin{array}{cc}\overline{)3}& \underset{¯}{\begin{array}{ccccc}2& -5& 7& -3& 1\\ & 6& & & \end{array}}\\ & \begin{array}{ccccc}2& & & & \end{array}\end{array}$ .

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

  $\begin{array}{cc}\overline{)3}& \underset{¯}{\begin{array}{ccccc}2& -5& 7& -3& 1\\ & 6& & & \end{array}}\\ & \begin{array}{ccccc}2& 1& & & \end{array}\end{array}$ .

Repeat the process until the end.

  $\begin{array}{cc}\overline{)3}& \underset{¯}{\begin{array}{ccccc}2& -5& 7& -3& 1\\ & 6& 3& 30& 81\end{array}}\\ & \begin{array}{ccccc}2& 1& 10& 27& 82\end{array}\end{array}$ .

Interpret:

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

The first four numbers ( $2$ , $1$ , $10$ and $27$ ) form the coefficients of the quotient, a third degree polynomial: $\left(2\right)x^3+\left(1\right)x^2+\left(10\right)x+(27)$ .

The last number $82$ is the remainder.

Thus, the quotient is $2x^3+x^2+10x+27$ and the remainder is $82$ .

Conclusion:

It is found that $2x^4-5x^3+7x^2-3x+1$ gives the quotient $2x^3+x^2+10x+27$ and the remainder $82$ when it is divided by $x-3$ .

Thus, it is found that:

  $\frac{2x^4-5x^3+7x^2-3x+1}{x-3}=\left(2x^3+x^2+10x+27\right)+\frac{82}{x-3}$ .