Chapter 2.3, Problem 1E

To determine

To identify : each term or function

Answer to Problem 1E

1st form of the Division Algorithm	2ndform of the Division Algorithm
$\underset{\stackrel{\uparrow }{{\displaystyle Dividend}}}{{\displaystyle f\left(x\right)}}\text{ = }\underset{\stackrel{\uparrow }{{\displaystyle Divisor}}}{{\displaystyle d\left(x\right)}}\cdot \underset{\stackrel{\uparrow }{{\displaystyle Quotient}}}{{\displaystyle q\left(x\right)}}+\underset{\stackrel{\uparrow }{{\displaystyle Remainder}}}{{\displaystyle r\left(x\right)}}$	$\frac{\stackrel{Dividend}{\overbrace{f\left(x\right)}}}{\underset{Divisor}{\underbrace{d\left(x\right)}}}=\stackrel{Quotient}{\overbrace{q\left(x\right)}}+\frac{\stackrel{\mathrm{Re}mainder}{\overbrace{r\left(x\right)}}}{\underset{Divisor}{\underbrace{d\left(x\right)}}}$

Explanation of Solution

Given information : Form (i) $f\left(x\right)=d\left(x\right)q\left(x\right)+r\left(x\right)$ ; (ii) $\frac{f\left(x\right)}{d\left(x\right)}=q\left(x\right)+\frac{r\left(x\right)}{d\left(x\right)}$

Concept Involved:

The Division Algorithm : If $f\left(x\right)$ and $d\left(x\right)$ are polynomials such that $d\left(x\right)\ne 0$ , and the degree of $d\left(x\right)$ isless than or equal to the degree of $f\left(x\right)$ , then there exist unique polynomials $q\left(x\right)$ and $r\left(x\right)$ such that $f\left(x\right)=d\left(x\right)\cdot q\left(x\right)+r\left(x\right)$ where $r\left(x\right)=0$ or the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ . If theremainder $r\left(x\right)$ is zero, then $d\left(x\right)$ divides evenly into $f\left(x\right)$ .

Another way to write the Division Algorithm is $\frac{f\left(x\right)}{d\left(x\right)}=q\left(x\right)+\frac{r\left(x\right)}{d\left(x\right)}$ .

In the Division Algorithm, the rational expression $f\left(x\right)/d\left(x\right)$ is improper because the degree of $f\left(x\right)$ is greater than or equal to the degree of $d\left(x\right)$ . On the other hand, the rational expression $r\left(x\right)/d\left(x\right)$ is proper because the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ .