Chapter 6.7, Problem 18WE

To determine

To divide: the expression $\frac{a^3+2a^2+3a+4}{a-1}$.

Answer to Problem 18WE

The quotient is $\left(a^2+3a+6\right)$ and the remainder is $10$.

Explanation of Solution

Given information:

  $\frac{a^3+2a^2+3a+4}{a-1}$

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 ,

  $\frac{a^3+2a^2+3a+4}{a-1}$

where

  $\begin{array}{l}\text{divisor}=a-1\\ \text{dividend}=a^3+2a^2+3a+4\end{array}$

Now, using polynomial division

Step 1.

  $a-1\begin{array}{l}\begin{array}{c}\hfill a^2\\ \hfill \overline{)\begin{array}{l}{a}^3+2a^2+3a+4\\ a^3-a^2\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $3a^2+3a+4$

Step 2.

  $a-1\begin{array}{l}\begin{array}{c}\hfill a^2+3a\\ \hfill \overline{)\begin{array}{l}3a^2+3a+4\\ 3a^2-3a\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $6a+4$

Step 3.

  $a-1\begin{array}{l}\begin{array}{c}\hfill a^2+3a+6\\ \hfill \overline{)\begin{array}{l}6a+4\\ 6a-6\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $10$

  $\frac{a^3+2a^2+3a+4}{a-1}=\left(a^2+3a+6\right)+\frac{10}{a-1}$

Thus, the quotient is $(a^2+3a+6)$ and the remainder is $10$.