# How should I solve this?

• May 3rd 2010, 07:41 PM
ChristopherDunn
How should I solve this?
Solve: $\displaystyle x^4-2x^3-2x-1=0$

I got the right solution, but I don't think I went about it the right way... I think I just had a lucky guess. How would you solve this?
• May 3rd 2010, 08:35 PM
TwistedOne151
hint
$\displaystyle x^4-2x^3-2x-1=x^4-2x^3-x^2+x^2-2x-1=(x^4-2x^3-x^2)+(x^2-2x-1)$
Can you proceed from here?

--Kevin C.
• May 3rd 2010, 09:15 PM
ChristopherDunn
Quote:

Originally Posted by TwistedOne151
$\displaystyle x^4-2x^3-2x-1=x^4-2x^3-x^2+x^2-2x-1=(x^4-2x^3-x^2)+(x^2-2x-1)$
Can you proceed from here?

--Kevin C.

How did you know which $\displaystyle x^2$ to pick?
• May 4th 2010, 12:53 PM
DeanSchlarbaum
Quote:

Originally Posted by TwistedOne151
$\displaystyle x^4-2x^3-2x-1=x^4-2x^3-x^2+x^2-2x-1=(x^4-2x^3-x^2)+(x^2-2x-1)$
Can you proceed from here?

--Kevin C.

Kevin: Is there some "method" you used to factor this equation? If so, would you mind showing me ALL (and perhaps others) the steps involved? Or, do you have that rare abilitiy where you just can look at an equation like the one Christopher listed and somehow "see" its factors?

I would appreciated it because I worked on this equation for quite a while and I could not figure out how exactly to factor it.
• May 4th 2010, 01:14 PM
bigwave
reorder the terms
Quote:

Originally Posted by ChristopherDunn
Solve: $\displaystyle x^4-2x^3-2x-1=0$

reorder the terms

$\displaystyle x^4-1-2x^3-2x=0$

then

$\displaystyle (x^4-1)-2x(x^2+1)=0$

$\displaystyle (x^2+1)(x^2-2x-1)=0$

$\displaystyle x=1-\sqrt{2}$
$\displaystyle x=1+\sqrt{2}$