Chapter 6.7, Problem 28WE

To determine

To divide: the expression $\frac{b^4+84}{b-3}$.

Answer to Problem 28WE

The quotient is $\left(b^3+3b^2+9b+27\right)$ and the remainder is $165$.

Explanation of Solution

Given information:

  $\frac{b^4+84}{b-3}$

Formula used:

  $\frac{\text{Dividend}}{\text{Divisor}}\text{=Quotient+}\frac{\text{Remainder}}{\text{Divisor}}$ .

Or

  $\text{Dividend=Quotient×Divisor+Remainder}$

Dividing the polynomials , each polynomial to be arranged in decreasing degree of the variable.

The missing terms in decreasing degree of variable to filled by using zero coefficients.

Calculation:

Consider ,

  $\frac{b^4+84}{b-3}$

where

  $\begin{array}{l}\text{divisor}=b-3\\ \text{dividend}=b^4+0b^3+0b^2++0b+84\end{array}$

Now, using polynomial division

Step 1.

  $b-3\begin{array}{l}\begin{array}{c}\hfill b^3\\ \hfill \overline{)\begin{array}{l}{b}^4+0b^3+0b^2+0b+84\\ b^4-3b^3\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $3b^3+0b^2+0b+84$

Step 2.

  $b-3\begin{array}{l}\begin{array}{c}\hfill b^3+3b^2\\ \hfill \overline{)\begin{array}{l}3b^3+0b^2+0b+84\\ 3b^3-9b^2\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $9b^2+0b+84$

Step 3.

  $b-3\begin{array}{l}\begin{array}{c}\hfill b^3+3b^2+9b\\ \hfill \overline{)\begin{array}{l}9b^2+0b+84\\ 9b^2-27b\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $27b+84$

Step 4.

  $b-3\begin{array}{l}\begin{array}{c}\hfill b^3+3b^2+9b+27\\ \hfill \overline{)\begin{array}{l}27b+84\\ 27b-81\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $165$

  $\frac{b^4+0b^3+0b^2+0b+84}{b-3}=\left(b^3+3b^2+9b+27\right)+\frac{165}{b-3}$

Thus, the quotient is $(b^3+3b^2+9b+27)$ and the remainder is $165$.