Chapter 11.3, Problem 26P

Divide.

$\left(3c^2-5c-2\right)÷\left(6c+2\right)$

To determine

To divide the polynomial

Answer to Problem 26P

The solution of the polynomial are $\frac{c}{2}-1$

Explanation of Solution

Given information:

The polynomial is $\left(3c^2-5c-2\right)÷\left(6c+2\right)$

Calculation:

The polynomial can be obtained as

  $\begin{array}{l}\left(3c^2-5c-2\right)÷\left(6c+2\right)\\ =\frac{\left(3c^2-5c-2\right)}{\left(6c+2\right)}\left[\text{dividing each term by }\left(6c+2\right)\right]\\ =\frac{3c^2-\left(6c-c\right)-2}{\left(6c+2\right)}\left[\text{write 5}c=\left(6c-c\right)\right]\\ =\frac{3c^2-6c+c-2}{\left(6c+2\right)}\\ =\frac{3c\left(c-2\right)+1\left(c-2\right)}{2\left(3c+1\right)}\left[\begin{array}{l}\text{the first and second term has a common factor,3}c\text{ and}\\ \text{the third and fourth term has a common factor,1 on the numerator}\end{array}\right]\\ =\frac{\left(c-2\right)\left(3c+1\right)}{2\left(3c+1\right)}\\ =\frac{\left(c-2\right)}{2}\left[\text{divide numerator and denominator by }\left(3c+1\right)\right]\\ =\frac{c}{2}-1\left[\text{divide numerator and denominator by }2\right]\end{array}$

Hence,

The solution of the polynomial are $\frac{c}{2}-1$