Chapter A.3, Problem 72AYU

To determine

To find: The quotient and the remainder and verify the answer.

Answer to Problem 72AYU

The quotient is $x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4$ and remainder is $0$ .

Explanation of Solution

Given information:

  $\text{Divide}\left(x^5-a^5\right)\text{by}\left(x-a\right)$ .

Calculation:

Divide the first term of dividend by the first term of divisor and use it as the first term of quotient.

  $x-a\begin{array}{c}\hfill x^4\\ \hfill \overline{)x^5-a^5}\end{array}$ .

Multiply $x^2$ by $x-a$

  $\begin{array}{l}=x^4\left(x-a\right)\\ =x^5-a{x}^4\end{array}$

Enter the result below the dividend.

  $x-a\begin{array}{c}\hfill x^4\\ \hfill \overline{)\begin{array}{l}{x}^5-a^5\\ x^5-a{x}^4\end{array}}\end{array}$ .

Subtract $x^5-a{x}^4$ from $x^5-a^5$ by changing sign of each term of $x^5-a{x}^4$ and write the remaining term below.

  $x-a\begin{array}{c}\hfill x^4\\ \hfill \overline{)\begin{array}{l}{x}^5-a^5\\ \underset{\_}{x^5-a{x}^4}\\ a{x}^4-a^5\end{array}}\end{array}$ .

Repeat the previous step by taking $a{x}^4-a^5$ .

  $x-a\begin{array}{c}\hfill x^4+a{x}^3\\ \hfill \overline{)\begin{array}{l}{x}^5-a^5\\ \underset{\_}{x^5-a{x}^4}\\ a{x}^4-a^5\\ \underset{\_}{a{x}^4-a^2{x}^3}\\ a^2{x}^3-a^5\\ \end{array}}\end{array}$ .

Further repeat the previous steps by taking $a^2{x}^3-a^5$ as dividend.

  $x-a\begin{array}{c}\hfill x^4+a{x}^3+a^2{x}^2\\ \hfill \overline{)\begin{array}{l}{x}^5-a^5\\ \underset{\_}{x^5-a{x}^4}\\ a{x}^4-a^5\\ \underset{\_}{a{x}^4-a^2{x}^3}\\ a^2{x}^3-a^5\\ \underset{\_}{a^2{x}^3-a^3{x}^2}\\ a^3{x}^2-a^5\\ \\ \end{array}}\end{array}$ .

Repeat the previous steps by taking $a^3{x}^2-a^5$ as dividend.

  $x-a\begin{array}{c}\hfill x^4+a{x}^3+a^2{x}^2+a^{3}x\\ \hfill \overline{)\begin{array}{l}{x}^5-a^5\\ \underset{\_}{x^5-a{x}^4}\\ a{x}^4-a^5\\ \underset{\_}{a{x}^4-a^2{x}^3}\\ a^2{x}^3-a^5\\ \underset{\_}{a^2{x}^3-a^3{x}^2}\\ a^3{x}^2-a^5\\ \underset{\_}{a^3{x}^2-a^{4}x}\\ a^{4}x-a^5\\ \\ \end{array}}\end{array}$ .

Repeat the previous steps by taking $a^{4}x-a^5$ as dividend.

The quotient is $x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4$ and remainder is $0$ .

To verify the answer write the relation between dividend, divisor, quotient and remainder

  $\left(\text{Quotient}\right)\times \left(\text{Divisor}\right)+\text{Remainder}=\text{Dividend}$ .

Put the values of quotient, divisor, remainder and dividend in the above equation and simplify

  $\begin{array}{c}\left(x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4\right)\times \left(x-a\right)+0=x^5-a^5\\ x\left(x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4\right)-a\left(x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4\right)=x^5-a^5\\ x^5+a{x}^4+a^2{x}^3+a^3{x}^2+a^{4}x-a{x}^4-a^2{x}^3-a^3{x}^2-a^{4}x-a^5=x^5-a^5\\ x^5-a^5=x^5-a^5\end{array}$ .

The result is verified

Therefore, the quotient is $x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4$ and remainder is $0$ .