Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Cool goIITian

Joined:	24 Aug 2008
Post:	30

9 Jul 2009 10:47:32 IST

0 People liked this

304

when u divide (x^4-6x^3+16x^2-25x+10) by (x^2-2x+k) , the remainder is x+a find a and k

None

when u divide (x^4-6x^3+16x^2-25x+10) by (x^2-2x+k) , the remainder is x+a find a and k

Share this article on:

Comments (3)

SAVVEJ

Blazing goIITian

Joined: 4 May 2007

Posts: 697

9 Jul 2009 11:10:27 IST

Like 1 people liked this

by simple division we get remainder as: (2k-9)x + (10- k(8-k))

equating with x+a we get

2k-9 = 1

k= 9/2

10 - k(8-k) = a

putting value of k we get,

a= -23/4

Siddharth Kothari

Blazing goIITian

Joined: 17 May 2009

Posts: 302

9 Jul 2009 11:23:49 IST

Like 0 people liked this

I agree with SAVVEJ

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

9 Jul 2009 11:48:40 IST

Like 1 people liked this

Here is a way to cut down the work..

x^2 equiv (2x-k) mod (x^2 -2x+k)

herefore x^4 - 6x^3+16x^2-25x+10 equiv (2x-k)^2-6(2x^2-kx)+16(2x-k)-25x+10

equiv -8x^2+x(2x+7)+(k^2-16k+10) equiv 8(k-2x)+x(2k+7)+(k^2-16k+10)

equiv x(2k-9) + (k^2-8k+10)

Now equate the co-efficiets ,

$2k-9=1\\Rightarrow k=5$

Also,

$k^2-8k+10=a\\Rightarrow a=-5$

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info

Find Posts by Topics

Physics.

Topics

9391

20087

3671

2186

7614

2826

2636

Mathematics.

Chemistry.

Biology

Institutes

Parents

Board

Fun Zone