• Oct 25th 2009, 12:34 PM
Matrim
I need to find the value of K in this soloution. I can't seem to figure it out.

x^2+(2k+1)x+(k^2-5)=0

Secondly, I need to factor

x^3-x^2-12x

First I use 3 as a factor and then I try and use synthetic division..
3 | 1 -1 -12 0
|___3__6 -18___
1 2 -6 -18
• Oct 25th 2009, 12:37 PM
artvandalay11
Quote:

Originally Posted by Matrim
I need to find the value of K in this soloution. I can't seem to figure it out.

x^2=(2k+1)x+(k2-5)=0

Secondly, I need to factor

x^3-x^2-12x

First I use 3 as a factor and then I try and use synthetic division..
3 | 1 -1 -12 0
|___3__6 -18___
1 2 -6 -18

Where are you getting stuck in the first one....

For the factoring question, you should first factor out an x, then you'll have a quadratic and won't need synthetic division
• Oct 25th 2009, 12:41 PM
Matrim
x^2+(2k+1)x+(k^2-5)=0

I typed that out wrong

i go..

x^2+2xk+x+k^2-5=0

We have been using the discriminant to do it, which numbers would i use?
• Oct 25th 2009, 12:44 PM
artvandalay11
Quote:

Originally Posted by Matrim
x^2+(2k+1)x+(k^2-5)=0

I typed that out wrong

i go..

x^2+2xk+x+k^2-5=0

We have been using the discriminant to do it, which numbers would i use?

Well if you wanna use the quadratic formula, you dont multiply it out,

\$\displaystyle ax^2+bx+c=0\$

In this case, \$\displaystyle b=(2k+1)\$, \$\displaystyle c=k^2-5\$, \$\displaystyle a=1\$

• Oct 25th 2009, 12:45 PM
Matrim
Quote:

Originally Posted by artvandalay11
Well if you wanna use the quadratic formula, you dont multiply it out,

\$\displaystyle ax^2+bx+c=0\$

In this case, \$\displaystyle b=(2k+1)\$, \$\displaystyle c=k^2-5\$, \$\displaystyle a=1\$