# Find the value of k using long division

• Oct 14th 2008, 01:11 PM
cjru
Find the value of k using long division
Hello guys, I'm really stuck with this question in my assignment. I just wondering if you can give me an idea how to solve K using long division. Here is the function.

F(x) = x^5 - 3x^4 + 7x^3 + Kx^2 + 9x - 5; d(x) = x^2 - 1 +1

What I did here is I divided F(x) by D(x). But I dont know what to do with the K.

• Oct 14th 2008, 03:00 PM
skeeter
Quote:

Originally Posted by cjru
Hello guys, I'm really stuck with this question in my assignment. I just wondering if you can give me an idea how to solve K using long division. Here is the function.

F(x) = x^5 - 3x^4 + 7x^3 + Kx^2 + 9x - 5; d(x) = x^2 - 1 +1

What I did here is I divided F(x) by D(x). But I dont know what to do with the K.

• Oct 14th 2008, 04:09 PM
cjru
Quote:

Originally Posted by skeeter

Sorry my divisor should be d(x) x^2-x+1
• Oct 14th 2008, 04:26 PM
cjru
Find the value of K using long division
My divisor should be d(x) = x^2 - x +1
• Oct 14th 2008, 04:31 PM
skeeter
I did the long division, and assuming the quotient has no remainder, i get k = -11.

Code:

```              x^3 - 2x^2 + 4x + (k+6) .............------------------------------------ x^2-x+1 | x^5 - 3x^4 + 7x^3 + kx^2 + 9x - 5               x^5 - x^4  + x^3 ..............------------------------------------                     -2x^4  + 6x^3 + kx^2 + 9x - 5                     -2x^4  + 2x^3 - 2x^2 ....................--------------------------------                                 4x^3 + (k+2)x^2 + 9x - 5                                 4x^3 -  4x^2    + 4x ...............................--------------------------                                           (k+6)x^2 + 5x - 5 ....................................... (k+6)x^2 - (k+6)x +(k+6)                                         -------------------------```
last line ...

k+6 = -5
k = -11
• Oct 14th 2008, 04:44 PM
cjru
Solving for K using a long division
Thank you very much for helping me.(Clapping) I just wanna clarify something. How did you get (k+2)x^2 .

2x^4 + 6x^3 + kx^2 + 9x - 5
-2x^4 + 2x^3 - 2x^2
--------------------------------
4x^3 + (k+2)x^2 + 9x - 5
4x^3 - 4x^2 + 4x
• Oct 14th 2008, 04:57 PM
skeeter
\$\displaystyle kx^2 - (-2x^2) = kx^2 + 2x^2 = (k+2)x^2\$
• Oct 14th 2008, 04:59 PM
cjru
Solving for K using a long division
On the last line

(k+6)x^2 + 5x-5
(k+6)x^2 - (k+6)x + (k+6)

how did you arrived this "k+6 = -5"

Thanks you for your patience. (Bow)
• Oct 14th 2008, 05:13 PM
skeeter
\$\displaystyle 5x - [-(k+6)x] = 0\$

\$\displaystyle 5x + (k + 6)x = 0\$

\$\displaystyle k+6 = -5\$
• Oct 14th 2008, 05:28 PM
cjru
Thank you very much
wow! Now everything is clear to me. Thank you very much Mr skeeter.(Clapping)

Godbless and keep up the good work. (Wink)