Advanced Functions - XP Math - Forums

 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

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