Chapter A.3, Problem 57AYU

To determine

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

Answer to Problem 57AYU

The quotient is $\left(4x^2-11x+23\right)$ and remainder is $-45$ .

Explanation of Solution

Given information:

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

Calculation:

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

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

Multiply $4x^2$ by $\left(x+2\right)$

  $\begin{array}{l}=4x^2\left(x+2\right)\\ =4x^3+8x^2\end{array}$

Enter the result below the dividend.

  $x+2\begin{array}{c}\hfill 4x^2\\ \hfill \overline{)\begin{array}{l}4x^3-3x^2+x+1\\ 4x^3+8x^2\end{array}}\end{array}$

Subtract $4x^3+8x^2$ from $4x^3-3x^2+x+1$ by changing sign of each term of $4x^3+8x^2$ and write the remaining term below.

  $x+2\begin{array}{c}\hfill 4x^2\\ \hfill \overline{)\begin{array}{l}4x^3-3x^2+x+1\\ \underset{\_}{4x^3+8x^2}\\ -11x^2+x+1\end{array}}\end{array}$ .

Repeat the previous step by taking $11x^2+x+1$ .

  $x+2\begin{array}{c}\hfill 4x^2-11x\\ \hfill \overline{)\begin{array}{l}4x^3-3x^2+x+1\\ \underset{\_}{4x^3+8x^2}\\ -11x^2+x+1\\ \underset{\_}{-11x^2-22x}\\ -23x+1\end{array}}\end{array}$ .

Further repeat the previous steps by taking $23x+1$ as dividend.

  $x+2\begin{array}{c}\hfill 4x^2-11x+23\\ \hfill \overline{)\begin{array}{l}4x^3-3x^2+x+1\\ \underset{\_}{4x^3+8x^2}\\ -11x^2+x+1\\ \underset{\_}{-11x^2-22x}\\ 23x+1\\ \underset{\_}{23x+46}\\ -45\end{array}}\end{array}$ .

Since further division is not possible.

The quotient is $\left(4x^2-11x+23\right)$ and remainder is $-45$ .

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

  $\begin{array}{c}\left(4x^2-11x+23\right)\times \left(x+2\right)-45=4x^3-3x^2+x+1\\ x\left(4x^2-11x+23\right)+2\left(4x^2-11x+23\right)-45=4x^3-3x^2+x+1\\ 4x^3-11x^2+23x+8x^2-22x+46-45=4x^3-3x^2+x+1\\ 4x^3-3x^2+x+1=4x^3-3x^2+x+1\end{array}$ .

The result is verified

Therefore, the quotient is $\left(4x^2-11x+23\right)$ and remainder is $-45$ .