Chapter 11.5, Problem 3BCYP

To determine

To calculate: To find the quotient $\left(n^2+6n+2\right)÷\left(n-2\right)$

Answer to Problem 3BCYP

The quotient is $n+8+\frac{18}{n-2}$

Explanation of Solution

Given information: Expression $\left(n^2+6n+2\right)÷\left(n-2\right)$

Formula Used:

Long division is the division of whole number in a systematic procedure to derive the answer.

Calculation:

Expression is given as

  $\left(n^2+6n+2\right)÷\left(n-2\right)$

Dividing the polynomial with the divisor to derive the quotient and remainder

  $n-2\begin{array}{c}\hfill n+8\\ \hfill \overline{)\begin{array}{l}{n}^2+6n+2\\ n^2-2n\\ -\text{ +}\\ -------\\ 0+8n+2\\ +8n-16\\ -\text{ +}\\ -------\\ \text{ 18}\end{array}}\end{array}$

Thus, quotient is

  $n+8+\frac{18}{n-2}$

Conclusion:

Hence, the quotient is $n+8+\frac{18}{n-2}$