Chapter A.3, Problem 65AYU

To determine

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

Answer to Problem 65AYU

The $\text{Quotient}=x^2-2x+\frac{1}{2}\text{and}\text{Remainder}=\frac{5}{2}x+\frac{1}{2}$ .

Explanation of Solution

Given information:

  $\text{Divide}\left(2x^4-3x^3+x+1\right)\text{by}\left(2x^2+x+1\right)$ .

Calculation:

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

  $2x^2+x+1\begin{array}{c}\hfill x^2\\ \hfill \overline{)2x^4-3x^3+x+1}\end{array}$ .

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

  $\begin{array}{l}=x^2\left(2x^2+x+1\right)\\ =2x^4+x^3+x^2\end{array}$

Enter the result below the dividend.

  $2x^2+x+1\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}2x^4-3x^3+x+1\\ 2x^4+x^3+x^2\end{array}}\end{array}$

Subtract $2x^4+x^3+x^2$ from $2x^4-3x^3+x+1$ by changing sign of each term of $2x^4+x^3+x^2$ and write the remaining term below.

  $2x^2+x+1\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}2x^4-3x^3+x+1\\ \underset{\_}{2x^4+x^3+x^2}\\ -4x^3-x^2+x+1\end{array}}\end{array}$ .

Repeat the previous steps by taking $-4x^3-x^2+x+1$ as dividend.

  $2x^2+x+1\begin{array}{c}\hfill x^2-2x\\ \hfill \overline{)\begin{array}{l}2x^4-3x^3+x+1\\ \underset{\_}{2x^4+x^3+x^2}\\ -4x^3-x^2+x+1\\ \underset{\_}{-4x^3-2x^2-2x}\\ x^2+3x+1\end{array}}\end{array}$ .

Further repeat the previous steps .

  $2x^2+x+1\begin{array}{c}\hfill x^2-2x+\frac{1}{2}\\ \hfill \overline{)\begin{array}{l}2x^4-3x^3+x+1\\ \underset{\_}{2x^4+x^3+x^2}\\ -4x^3-x^2+x+1\\ \underset{\_}{-4x^3-2x^2-2x}\\ x^2+3x+1\\ \underset{\_}{x^2+\frac{1}{2}x+\frac{1}{2}}\\ \frac{5}{2}x+\frac{1}{2}\end{array}}\end{array}$

Since further division is not possible $\text{Quotient}=x^2-2x+\frac{1}{2}\text{and}\text{Remainder}=\frac{5}{2}x+\frac{1}{2}$ .

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

  $\begin{array}{c}\left(x^2-2x+\frac{1}{2}\right)\times \left(2x^2+x+1\right)+\left(\frac{5}{2}x+\frac{1}{2}\right)=2x^4-3x^3+x+1\\ x^2\left(2x^2+x+1\right)-2x\left(2x^2+x+1\right)+\frac{1}{2}\left(2x^2+x+1\right)+\left(\frac{5}{2}x+\frac{1}{2}\right)=2x^4-3x^3+x+1\\ 2x^4+x^3+x^2-4x^3-2x^2-2x+x^2+\frac{1}{2}x+\frac{1}{2}+\frac{5}{2}x+\frac{1}{2}=2x^4-3x^3+x+1\\ 2x^4+\left(x^3-4x^3\right)+\left(-2x+\frac{1}{2}x+\frac{5}{2}x\right)+\left(\frac{1}{2}+\frac{1}{2}\right)=2x^4-3x^3+x+1\end{array}$ .

  $2x^4-3x^3+x+1=2x^4-3x^3+x+1$

Thus, it is verified .

Therefore, the $\text{Quotient}=x^2-2x+\frac{1}{2}\text{and}\text{Remainder}=\frac{5}{2}x+\frac{1}{2}$ .