Chapter A.3, Problem 64AYU

To determine

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

Answer to Problem 64AYU

The $\text{Quotient}=x^2\text{and}\text{Remainder}=x-2$

Explanation of Solution

Given information:

  $\text{Divide}\left(3x^5-x^2+x-2\right)\text{by}\left(3x^3-1\right)$ .

Calculation:

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

  $3x^3-1\begin{array}{c}\hfill x^2\\ \hfill \overline{)3x^5-x^2+x-2}\end{array}$ .

Multiply $x^2$ by $\left(3x^3-1\right)$ .

  $\begin{array}{l}=x^2\left(3x^3-1\right)\\ =3x^5-x^2\end{array}$

Enter the result below the dividend.

  $3x^3-1\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}3x^5-x^2+x-2\\ 3x^5-x^2\end{array}}\end{array}$

Subtract $3x^5-x^2$ from $3x^5-x^2+x-2$ by changing sign of each term of $3x^5-x^2$ and write the remaining term below.

  $3x^3-1\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}3x^5-x^2+x-2\\ \underset{\_}{3x^5-x^2}\\ x-2\end{array}}\end{array}$ .

Since further division is not possible $\text{Quotient}=x^2\text{and}\text{Remainder}=x-2$ .

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

  $\begin{array}{c}\left(x^2\right)\times \left(3x^3-1\right)+\left(x-2\right)=3x^5-3x^2+x-2\\ 3x^5-3x^2+x-2=3x^5-3x^2+x-2\end{array}$ .

Thus, it is verified.

Therefore, the $\text{Quotient}=x^2\text{and}\text{Remainder}=x-2$ .