- 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
Find the remainder when $x^3+ 3x^2 + 3x + 1$ is divided by
\( x-\frac{1}{2} \)
Given:
$x^3+ 3x^2 + 3x + 1$ is divided by $x-\frac{1}{2}$
To do:
We have to find the remainder when $x^3+ 3x^2 + 3x + 1$ is divided by $x-\frac{1}{2}$.
Solution:
The remainder theorem states that when a polynomial, $p(x)$ is divided by a linear polynomial, $x - a$ the remainder of that division will be equivalent to $p(a)$.
Let $f(x) = x^3+ 3x^2 + 3x + 1$ and $g(x) = x-\frac{1}{2}$
So, the remainder will be $f(\frac{1}{2})$.
$f(\frac{1}{2}) = (\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2}) + 1$
$= \frac{1}{8} + 3(\frac{1}{4}) +\frac{3}{2}+1$
$=\frac{1+3(2)+3(4)+1(8)}{8}$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$
Therefore, the remainder is $\frac{27}{8}$.
Advertisements