Factorising Cubics

Nov 2010
17
0
I have asked way too many questions, and Im really sorry if I have bothered you lot too much, but thanks so much for the help I do apprieceate every single comment, and it is a great help to my studies. (Worried)

Just a question regarding cubics, I need to factorise
x^3-kx^2+2kx-k-1, I already know x-1 is a factor. If I want to factorise this via the "long division method" how do you go about in doing that? I'm just confused with all these pronumerals, thanks for the help.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Let's call this function \(\displaystyle \displaystyle P(x) = x^3 - kx^2 + 2kx - k - 1 = 0\).

You can start by evaluating \(\displaystyle \displaystyle k\). Since you know \(\displaystyle \displaystyle x - 1\) is a factor, that means \(\displaystyle \displaystyle P(1) = 0\).
 
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Nov 2010
17
0
Oh, I am such an idiot ;). Its the case of rewriting the equation and solving for k correct?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Yes, once you have substituted x = 1 and noted that this is a value that makes the entire polynomial = 0.
 
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Nov 2010
17
0
Yeah, I have subbed x=1 one in, and ended up with 0. Now that I know x-1 is a factor do I go ahead with the long division?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Oops, I didn't realise that substituting x = 1 just gives 0 = 0. You can't evaluate k this way.

When you are doing the long division, you need to treat -k-1 as a single term, and remember that since x-1 is a factor, the remainder will be 0.
 
Oct 2010
115
17
Heart of Winter
\left[ \right] you should try the " "long division method" " on this expression and post you work so we can help.