# Factoring

• September 22nd 2009, 10:58 PM
jzellt
Factoring
How do you factor x^3 - 2x - 1 ?
• September 22nd 2009, 11:04 PM
Matt Westwood
Trial and error. In this case it looks fairly straightforward.

If you're going to have a factor, there's going to be either x+1 or x-1, because x times something needs to make $x^3$ and you've got a $-1$ at the end. So you know that the first thing you do is try and work out how to make $x^3 - 2x -1$ like $(x \pm 1) (x^2 +ax \pm 1)$.

What I would do would be to try and extract $x+1$ and then if that doesn't work $x-1$ by long division.
• September 22nd 2009, 11:07 PM
mr fantastic
Quote:

Originally Posted by jzellt
How do you factor x^3 - 2x - 1 ?

Note that x = -1 is a root and therefore (x + 1) is a factor. Now divide (x + 1) into the cubic to get the (irreducible) quadratic factor.
• September 22nd 2009, 11:15 PM
jzellt
Can you show me the steps? I really have forgotten this stuff
• September 23rd 2009, 01:13 AM
Quote:

Originally Posted by jzellt
How do you factor x^3 - 2x - 1 ?

HI

List down the factors of the constant term of this equation and test whether they are the zeros of the cubic expression .

ie 1 , -1

$f(1)=1^3-2(1)-1=-2$ so $(x-1)$ is not a factor .

$f(-1)=(-1)^3-2(-1)-1=0$ so $(x+1)$ is a factor .

And as suggested by Mr F , do the long division to find the quadratic factor which is complex .
• September 23rd 2009, 03:34 AM
CaptainBlack
Quote:

Originally Posted by jzellt
Can you show me the steps? I really have forgotten this stuff

If you do not know how to do synthetic division or polynomial long division (and Wikipedia has an article explaing this if you really have forgotten) you can do this by inspection since you have:

$(x+1)(x^2+bx+c)=x^3-2x-1$

expand the product on the left and equate the coefficients on either side of corresponding powers of $x$ to solve for $b$ and $c$

CB
• September 23rd 2009, 06:36 AM
stapel
Quote:

Originally Posted by jzellt
Can you show me the steps? I really have forgotten this stuff

To re-learn the steps for factoring polynomials, try here. Make sure to review the Rational Roots Test and synthetic division, if necessary. (Wink)
• September 24th 2009, 05:34 AM
pacman
x^3 - 2x - 1 = 0?