Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Divide the folowing:$ 9 l^{3}+18 l^{2}-9$ by $3 l+3$
Given :
The given terms are $ 9 l^{3}+18 l^{2}-9$ and $3 l+3$.
To do :
We have to divide $ 9 l^{3}+18 l^{2}-9$ by $3 l+3$.
Solution :
Dividend $= 9 l^{3}+18 l^{2}-9$
Divisor $= 3l+3$
$3l^2+3l-3$
________________
$3l+3 | 9 l^{3}+18 l^{2}-9$ $\frac{9l^3}{3l} = 3l^2$
| $9l^3 + 9l^2$
---------------
$9l^2 -9$ $\frac{9l^2}{3l} = 3l$
$9l^2 +9l$
---------
$-9l-9$ $\frac{-9l}{3l} = -3$
$-9l-9$
---------
0
Quotient $= 3l^2+3l-3$
Remainder$ = 0$
Verification:
$Dividend = Divisor \times Quotient + Remainder$
$ = (3l+3)( 3l^2+3l-3 )+0$
$ = 3l( 3l^2+3l-3) +3( 3l^2+3l-3 )$
$= 9l^3+9l^2-9l+9l^2+9l-9$
$= 9l^3+18l^2-9$
Hence verified.
35 Views
