# (Year 10) Factoring 6th Degree Polynomial

• Oct 1st 2008, 05:52 AM
essaymasters
(Year 10) Factoring 6th Degree Polynomial
Hey everyone! I'm new here and I'd be grateful of your help.

I would like to know how to factor and find the roots/zeroes of:
$14x^6-9x^2+2$

I heard of a method where you can just substitute $a=x^2$ but I couldn't get any steps past it: $14x^3-9x+2$

Thanks for your time!

--essaymasters
• Oct 1st 2008, 10:45 AM
Laurent
Hi,
Did you find this equation in an homework? The trick with $y=x^2$ is a good one, but the equation $14y^3-9y+2=0$ has no "easy" roots. You can use Cardano's method for instance, which works for any cubic equation. However, you'll probably have to deal with cubic roots of complex numbers, while the three roots are in fact real. I asked Maple (a computer program) for solutions, and for instance one of them can be written as (after manual simplifications of maple's output): $-\sqrt{\frac{6}{7}}\cos\left(\frac{1}{3}\arctan\sqr t{\frac{13}{14}}\right)\simeq -0.8957$... If this is an homework, I'd bet you made a mistake before. If it is a "real life" problem, use approximate solutions: -0.8957, 0.245, 0.650 for $14y^3-9y+2=0$ and hence the six square roots (with $\pm$) as solutions of the initial equation in $\mathbb{C}$, and only four solutions (plus or minus the roots of the two positive roots) in $\mathbb{R}$.
• Oct 1st 2008, 03:07 PM
essaymasters
Thanks Laurent, yes this was a homework problem but since no-one including the teach could factor it, it may have been a typo, especially considering our unit.

(Rock)