Quantcast Advanced Functions - XP Math - Forums
XP Math Home Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment PDF Version

Go Back   XP Math - Forums > Mathematics > Homework Help

Reply
 
Thread Tools Display Modes
Old 07-07-2009   #1
cirquedufreak
 
Join Date: Jul 2009
Posts: 1
Default Advanced Functions

Could Someone please help me...I have no idea what to do for these two promblems

For what values of k does the function f(x) = x3 + 6x2 + kx 4 give the same remainder when divided by (x-1) and (x + 2)?

Determine the value of k when x3 + kx2 + 2x 3 is divided by x + 2, and the remainder is 1.
cirquedufreak is offline   Reply With Quote
Old 07-09-2009   #2
MAS1

Points: 8,254, Level: 61
Points: 8,254, Level: 61 Points: 8,254, Level: 61 Points: 8,254, Level: 61
Activity: 33.3%
Activity: 33.3% Activity: 33.3% Activity: 33.3%
Last Achievements
 
Join Date: Dec 2008
Posts: 249
Default

Quote:
Originally Posted by cirquedufreak View Post
Could Someone please help me...I have no idea what to do for these two promblems

For what values of k does the function f(x) = x3 + 6x2 + kx 4 give the same remainder when divided by (x-1) and (x + 2)?

Determine the value of k when x3 + kx2 + 2x 3 is divided by x + 2, and the remainder is 1.
1. Divide x^3 + 6x^2 + kx - 4 by (x - 1). This gives x^2 + 7x + (k + 7) with a remainder of k + 3. Then divide x^3 + 6x^2 + kx - 4 by (x + 2).This gives x^2 + 4x + (k - 8) with a remainder of 12 - 2k. Then set the two remainders equal to each other.

k + 3 = 12 - 2k
3k = 9
k = 3

2. Divide x^3 + kx^2 + 2x - 3 by x + 2 giving x^2 + (k -2)x + (2 - 2(k - 2)) with a remainder of -3 - 2(2 - 2(k -2)). Then set the remainder equal to 1 and solve for k.

-3 - 2(2 - 2(k - 2)) = 1
-2(2 - 2k + 4) = 4
2 - 2k + 4 = -2
6 - 2k = -2
-2k = -8
k = 4
MAS1 is offline   Reply With Quote
Old 02-22-2010   #3
jmw106462

Last Achievements
 
jmw106462's Avatar
 
Join Date: Feb 2010
Posts: 388
Default

1. Divide x^3 + 6x^2 + kx - 4 by (x - 1). This gives x^2 + 7x + (k + 7) with a remainder of k + 3. Then divide x^3 + 6x^2 + kx - 4 by (x + 2).This gives x^2 + 4x + (k - 8) with a remainder of 12 - 2k. Then set the two remainders equal to each other.

k + 3 = 12 - 2k
3k = 9
k = 3

2. Divide x^3 + kx^2 + 2x - 3 by x + 2 giving x^2 + (k -2)x + (2 - 2(k - 2)) with a remainder of -3 - 2(2 - 2(k -2)). Then set the remainder equal to 1 and solve for k.

-3 - 2(2 - 2(k - 2)) = 1
-2(2 - 2k + 4) = 4
2 - 2k + 4 = -2
6 - 2k = -2
-2k = -8
k = 4
jmw106462 is offline   Reply With Quote
Old 02-24-2010   #4
MAS1

Points: 8,254, Level: 61
Points: 8,254, Level: 61 Points: 8,254, Level: 61 Points: 8,254, Level: 61
Activity: 33.3%
Activity: 33.3% Activity: 33.3% Activity: 33.3%
Last Achievements
 
Join Date: Dec 2008
Posts: 249
Default Cut and Paste?

Quote:
Originally Posted by jmw106462 View Post
1. Divide x^3 + 6x^2 + kx - 4 by (x - 1). This gives x^2 + 7x + (k + 7) with a remainder of k + 3. Then divide x^3 + 6x^2 + kx - 4 by (x + 2).This gives x^2 + 4x + (k - 8) with a remainder of 12 - 2k. Then set the two remainders equal to each other.

k + 3 = 12 - 2k
3k = 9
k = 3

2. Divide x^3 + kx^2 + 2x - 3 by x + 2 giving x^2 + (k -2)x + (2 - 2(k - 2)) with a remainder of -3 - 2(2 - 2(k -2)). Then set the remainder equal to 1 and solve for k.

-3 - 2(2 - 2(k - 2)) = 1
-2(2 - 2k + 4) = 4
2 - 2k + 4 = -2
6 - 2k = -2
-2k = -8
k = 4
Nice job of cutting and pasting!
MAS1 is offline   Reply With Quote
Old 02-24-2010   #5
jmw106462

Last Achievements
 
jmw106462's Avatar
 
Join Date: Feb 2010
Posts: 388
Default

Lol i was seeing if u would catch that you did.*of) Course, The other post i didn't Cut and Paste Thou
__________________
Please refrain from sending me frivolous PM's
jmw106462 is offline   Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 06:52 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2017, vBulletin Solutions Inc.
XP Math