Chapter 11.3, Problem 39PPE

To determine

To divide $\left(5t^4-10t^2+6\right)÷\left(t+5\right)$ .

Answer to Problem 39PPE

Dividing $\left(5t^4-10t^2+6\right)÷\left(t+5\right)$ the quotient is $5t^3-25t^2+115t-575$ and remainder is 2869.

Explanation of Solution

Given:

  $\left(5t^4-10t^2+6\right)÷\left(t+5\right)$

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 $\left(5t^4-10t^2+6\right)÷\left(t+5\right)$ use long division method.

Since the terms of the dividend and divisor in are already in descending powers of the variable,

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

  $\left(5t^4+0t^3-10t^2+0t+6\right)÷\left(t+5\right)$

Thus, the given division may be proceeded as:

  $\begin{array}{l}t+5)5t^4+0t^3-10t^2+0t+6(5t^3-25t^2+115t-575\\ \underset{\_}{5t^4+25t^3}\\ \underset{\_}{\begin{array}{l}-25t^3-10t^2+0t+6\\ -25t^3-125t^2\end{array}}\\ 115t^2+0t+6\\ \underset{\_}{115t^2+575t}\\ -575t+6\\ \underset{\_}{-575t-2875}\\ \text{ 2869}\end{array}$

Thus, dividing $\left(5t^4-10t^2+6\right)÷\left(t+5\right)$ the quotient is $5t^3-25t^2+115t-575$ and remainder is 2869.