Chapter 6.8, Problem 18SGR

To determine

To calculate: The expression,

  $\frac{6y^3+13y^2-10y-24}{y+2}$

Answer to Problem 18SGR

The factor of solution of the expression is $\left(y+2\right)\left(2y+3\right)\left(3y-4\right)$ .

Explanation of Solution

Given information:

The expression is given as:

  $\frac{6y^3+13y^2-10y-24}{y+2}$

Formula used:

The process when divide a polynomial by a polynomial with more than one term is known as division algorithm.

Calculation:

Consider the expression,

  $\frac{6y^3+13y^2-10y-24}{y+2}$

Using the long division method,

  $y+2\overline{)6y^3+13y^2-10y-24}$

Multiply divisor by $6y^3$ since $\frac{6y^3}{6y^2}=y$ .

Subtract and bring down the next term as,

  $\begin{array}{l}y+2\begin{array}{c}\hfill 6y^2\\ \hfill \overline{)\begin{array}{l}6y^3+13y^2-10y-24\\ 6y^3+12y^2\\ --\\ \overline{y^2-10y-24}\end{array}}\end{array}\\ \end{array}$

Now, multiply divisor by $y$ since, $\frac{y^2}{y}=y$ .

Subtract and bring down the next term as,

  $\begin{array}{l}y+2\begin{array}{c}\hfill 6y^2+y\\ \hfill \overline{)\begin{array}{l}6y^3+13y^2-10y-24\\ 6y^3+12y^2\\ --\\ \overline{\begin{array}{l}{y}^2-10y-24\\ y^2+2y\\ \overline{\text{}\text{}\text{}\text{}\text{}\text{}-12y-24}\end{array}}\end{array}}\end{array}\\ \end{array}$

Now, multiply divisor by $y$ since, $\frac{-12y}{y}=-12$ .

Subtract and bring down the next term as,

  $\begin{array}{l}y+2\begin{array}{c}\hfill 6y^2+y-12\\ \hfill \overline{)\begin{array}{l}6y^3+13y^2-10y-24\\ 6y^3+12y^2\\ --\\ \overline{\begin{array}{l}{y}^2-10y-24\\ y^2+2y\\ \overline{\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}-12y-24\\ -12y-24\\ \overline{0}\end{array}}\end{array}}\end{array}}\end{array}\\ \end{array}$

The quotient is $6y^2+y-12$ and remainder is 0. The polynomial $6y^2+y-12$ is factorized as,

  $\begin{array}{l}6y^2+y-12\\ 6y^2+9y-8y-12\\ 3y\left(2y+3\right)-4\left(2y+3\right)\\ \left(2y+3\right)\left(3y-4\right)\end{array}$

Therefore, the factor is $\left(y+2\right)\left(2y+3\right)\left(3y-4\right)$ .

Thus, the factor of solution of the expression is $\left(y+2\right)\left(2y+3\right)\left(3y-4\right)$