Chapter 2.4, Problem 11E

Solution

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

It is found that:

  $\frac{5x^4-3x+1}{4-x}=\left(-5x^3-20x^2-80x-317\right)+\frac{1269}{4-x}$ .

That is, $5x^4-3x+1$ gives the quotient $-5x^3-20x^2-80x-317$ and the remainder $1269$ when it is divided by $4-x$ .

Given:

The dividend:

  $5x^4-3x+1$

The divisor:

  $4-x$

Set up:

Observe that $\frac{5x^4-3x+1}{4-x}=\frac{-5x^4+3x-1}{x-4}$ .

Thus, the problem can be viewed as that of dividing $-5x^4+3x-1$ by $x-4$ .

Now, the zero of the divisor $x-4$ is $4$ . So, put $4$ in the divisor position.

The dividend $-5x^4+3x-1$ is not in the standard form.

Thus, write it in the standard form at first:

  $-5x^4+0x^3+0x^2+3x-1$ .

Now, write the coefficients of the dividend, in order, in the dividend position.

  $\begin{array}{cc}\overline{)4}& \underset{¯}{\begin{array}{ccccc}-5& 0& 0& 3& -1\\ & & & & \end{array}}\\ & \begin{array}{ccccc}& & & & \end{array}\end{array}$ .

Calculation:

Because the leading coefficient of the dividend must be the leading coefficient of the quotient, write $-5$ in the first quotient position.

  $\begin{array}{cc}\overline{)4}& \underset{¯}{\begin{array}{ccccc}-5& 0& 0& 3& -1\\ & & & & \end{array}}\\ & \begin{array}{ccccc}-5& & & & \end{array}\end{array}$ .

Multiply the zero of the divisor, $4$ , by the most recently determined coefficient of the quotient , $-5$ . Now, write this product above the line and one column to the right.

  $\begin{array}{cc}\overline{)4}& \underset{¯}{\begin{array}{ccccc}-5& 0& 0& 3& -1\\ & -20& & & \end{array}}\\ & \begin{array}{ccccc}-5& & & & \end{array}\end{array}$ .

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

  $\begin{array}{cc}\overline{)4}& \underset{¯}{\begin{array}{ccccc}-5& 0& 0& 3& -1\\ & -20& & & \end{array}}\\ & \begin{array}{ccccc}-5& -20& & & \end{array}\end{array}$ .

Repeat the process until the end.

  $\begin{array}{cc}\overline{)4}& \underset{¯}{\begin{array}{ccccc}-5& 0& 0& 3& -1\\ & -20& -80& -320& -1268\end{array}}\\ & \begin{array}{ccccc}-5& -20& -80& -317& -1269\end{array}\end{array}$ .

Interpret:

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

The first four numbers ( $-5$ , $-20$ , $-80$ and $-317$ ) form the coefficients of the quotient, a third degree polynomial: $\left(-5\right)x^3+\left(-20\right)x^2+\left(-80\right)x+(-317)$ .

The last number $-1269$ is the remainder.

Thus, the quotient is $-5x^3-20x^2-80x-317$ and the remainder is $-1269$ .

Conclusion:

It is found that $-5x^4+3x-1$ gives the quotient $-5x^3-20x^2-80x-317$ and the remainder $-1269$ when it is divided by $x-4$ .

Thus, it is found that:

  $\frac{-5x^4+3x-1}{x-4}=\left(-5x^3-20x^2-80x-317\right)-\frac{1269}{x-4}$ .

That is, it is found that:

  $\frac{5x^4-3x+1}{4-x}=\left(-5x^3-20x^2-80x-317\right)+\frac{1269}{4-x}$ .

That is, $5x^4-3x+1$ gives the quotient $-5x^3-20x^2-80x-317$ and the remainder $1269$ when it is divided by $4-x$ .