How do you factor x^3 - 2x - 1 ?

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- Sep 22nd 2009, 10:58 PMjzelltFactoring
How do you factor x^3 - 2x - 1 ?

- Sep 22nd 2009, 11:04 PMMatt 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 and you've got a at the end. So you know that the first thing you do is try and work out how to make like .

What I would do would be to try and extract and then if that doesn't work by long division. - Sep 22nd 2009, 11:07 PMmr fantastic
- Sep 22nd 2009, 11:15 PMjzellt
Can you show me the steps? I really have forgotten this stuff

- Sep 23rd 2009, 01:13 AMmathaddict
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

so is not a factor .

so is a factor .

And as suggested by Mr F , do the long division to find the quadratic factor which is complex . - Sep 23rd 2009, 03:34 AMCaptainBlack
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:

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

CB - Sep 23rd 2009, 06:36 AMstapel
To re-learn the steps for factoring polynomials, try

**here**. Make sure to review**the Rational Roots Test**and**synthetic division**, if necessary. (Wink) - Sep 24th 2009, 05:34 AMpacman
x^3 - 2x - 1 = 0?

add and subtract x^2

x^3 + x^2 - x^2 - 2x - 1, rearrange,

(x^3 + x^2) +(-x^2 - 2x - 1) = 0, factor again

x^2(x + 1) - (x + 1)(x + 1) = 0, factor (x +1)

(x + 1)(x^2 - x - 1) = ?