Chapter 11.3, Problem 35PPE

To determine

To divide $\left(4y+9y^3-7\right)÷\left(-5+3y\right)$ .

Answer to Problem 35PPE

Dividing $\left(4y+9y^3-7\right)÷\left(-5+3y\right)$ the quotient is $3y^2+5y+\frac{29}{3}$ and remainder is $\frac{8}{3}$ .

Explanation of Solution

Given:

  $\left(4y+9y^3-7\right)÷\left(-5+3y\right)$

Concept Used:

Steps to follow before applying Long Division method:

1. Write the terms of the dividend and divisor in descending powers of the variable.
2. Insert placeholders with zero coefficients for missing powers of the variable.

Calculation:

To divide $\left(4y+9y^3-7\right)÷\left(-5+3y\right)$ use long division method.

First, arrange the terms of the dividend and divisor in descending powers of the variable, as

  $\left(9y^3+4y-7\right)÷\left(3y-5\right)$

Now, insert placeholders with zero coefficients for missing powers of the variable, as

  $\left(9y^3+0y^2+4y-7\right)÷\left(3y-5\right)$

Thus, the given division may be proceeded as:

  $\begin{array}{l}3y-5)9y^3+0y^2+4y-7(3y^2+5y+\frac{29}{3}\\ \underset{\_}{9y^3-15y^2}\\ \underset{\_}{\begin{array}{l}15y^2+4y\\ 15y^2-25y\end{array}}\\ 29y-7\\ \underset{\_}{29y-\frac{29}{3}}\\ \frac{29}{3}-7=\frac{29-21}{3}\\ \text{And }\frac{29-21}{3}=\frac{8}{3}\end{array}$

Thus, dividing $\left(4y+9y^3-7\right)÷\left(-5+3y\right)$ the quotient is $3y^2+5y+\frac{29}{3}$ and remainder is $\frac{8}{3}$ .