 Advanced Functions - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Advanced Functions 07-07-2009 #1 cirquedufreak   Join Date: Jul 2009 Posts: 1 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.  07-09-2009   #2
MAS1   Join Date: Dec 2008
Posts: 249 Quote:
 Originally Posted by cirquedufreak 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  02-22-2010 #3 jmw106462      Join Date: Feb 2010 Posts: 388 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  02-24-2010   #4
MAS1   Join Date: Dec 2008
Posts: 249 Cut and Paste?

Quote:
 Originally Posted by jmw106462 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!  02-24-2010 #5 jmw106462      Join Date: Feb 2010 Posts: 388 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 Thread Tools Show Printable Version Email this Page Display Modes Switch to Linear Mode Hybrid Mode Switch to Threaded Mode 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 Rules
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