Vriti Edustore
Home »
Ask & Discuss
»
Mathematics
« Back to Discussion
Algebra
Comments (3)
Ken kronser
New kid on the Block
Joined: 1 Feb 2011
Posts: 2
1 Feb 2011 06:53:05 IST
Like
0 people liked this
Duh
Reply
1 Feb 2011 14:56:29 IST
Like
0 people liked this
By trial we can see 1 is a zero of the polynomial P(x) = x3 - 10x2 + 29x - 20
So, either divide it by (x - 1) or write in such a way so that x - 1 comes out to be common.
As, P(x) = x3 - x2 - 9x2 + 9x + 20x - 20 = x2 (x - 1) - 9x (x - 1) + 20 (x - 1) = (x - 1)(x2 - 9x + 20)
= (x - 1)(x2 - 5x - 4x + 20) = (x - 1)[x(x - 5) - 4(x - 5)] = (x - 1)(x - 5)(x - 4)
Or, P(x) = (x - 1)(x - 4)(x - 5)
Free Sign Up!
|
|
|
|
Preparing for JEE?
Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015
@ INR 5,443/-
For Quick Info
Vertical Limit 
Top Contributors
All Time This Month Last Week | 1. |
|
Bipin Dubey
|
| 2. |
|
Himanshu
|
| 3. |
|
Hari Shankar
|
| 4. |
|
edison
|
| 5. |
|
Sagar Saxena
|
| 6. |
|
Yagyadutt Mishr..
|
Find Posts by Topics
Physics
Topics
10300
21014
3875
2311
8099
2973
2797
Mathematics
Chemistry
Biology
Institutes
Parents Corner
Board
Fun Zone
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1223 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011