Posted by **BioBabe** on Sunday, October 12, 2008 at 11:48pm.

How would you factor 3x^3-4x^2+4x-1?

P.S. Factor theorem does not work here.

- 12th grade Math? -
**drwls**, Monday, October 13, 2008 at 2:34am
Look for the first root by trial and error. Using the "q/p theorem", I see that one solution is +1/3. So, (x - 1/3) is one factor.

Divide that into 3x^3-4x^2+4x-1 to get a quadratic factor,

(3x^2 -3x +3) = 3 (x^2 -x +1).

The term in parentheses can be factored by the usual means, but gives two complex roots.

3x^3-4x^2+4x-1 = 3(x- 1/3)(x^2-x+1)

(3x-1)(x^2-x+1))

- 12th grade Math? -
**BioBabe**, Monday, October 13, 2008 at 11:03am
What's the q/p theorem though?

- 12th grade Math? -
**drwls**, Monday, October 13, 2008 at 12:03pm
I was afraid you'd ask me that :-)

The more common name for it is the rational roots test (or theorem).

Here is a reference.

http://en.wikipedia.org/wiki/Rational_root_theorem

It doesn't always provide a root, but if there is a rational real root, it works.

Briefly, it says that if the constant term of the polynomial is q and the first term in p, and if there are rational real roots, one of the roots will be

+/- q/p or +/- the tio of prime-number factors of q and p.

- eapmo dzywoakr -
**eapmo dzywoakr**, Friday, February 6, 2009 at 12:26pm
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