Chapter 6, Problem 8MCQ

To determine

To calculate: The simplified form of the expression $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)$ .

Answer to Problem 8MCQ

The simplified form of the expression $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)$ is $\left(2r^2-r-9\right)+\frac{-25}{\left(2r-5\right)}$ .

Explanation of Solution

Given information:

The expression is $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)$ .

Formula used:

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Calculation:

Consider the expression $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)$ .

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Apply the method of long division. First multiply the expression $2r-5$ by $2r^2$ . Subtract it from the dividend. Next multiply $2r-5$ by $-r$ . Subtract the expression obtained. Again multiply 2r-5 by -9. Subtract the expression obtained.

  $\begin{array}{l}2r-5\begin{array}{c}\hfill 2r^2-r-9\\ \hfill \overline{)4r^3-8r^2-13r+20}\end{array}\\ \begin{array}{ccc}& & 4r^3-10r^2\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& 2r^2-13r\end{array}\end{array}\\ \begin{array}{cc}& \begin{array}{ccc}& & 2r^2+5r\end{array}\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& 0-18r\end{array}\end{array}+20\\ \begin{array}{cc}& \begin{array}{ccc}& & 0-18r\end{array}\end{array}+45\\ \begin{array}{cc}& \begin{array}{cccc}\begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}& & -25& \end{array}\end{array}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ . Here, dividend is $4r^3-8r^2-13r-9$ quotient is $2r^2-r-9$ , divisor is $2r-5$ and remainder is $-25$ .

Therefore, $4r^3-8r-13r+20=\left(2r-5\right)\left(2r^2-r-9\right)+\left(-25\right)$ that is,

  $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)=\left(2r^2-r-9\right)+\frac{-25}{\left(2r-5\right)}$

Thus, the simplified form of the expression $\left(4r^3-8r^2-13r+20\right)÷\left(2r-5\right)$ is $\left(2r^2-r-9\right)+\frac{-25}{\left(2r-5\right)}$ .