Roots of the polynomial

[wishmaster]

I have given polynomial: \(\displaystyle x^3-8x^2+19x-12\)

I know how to find the roots with Horners method,i am just wondering if there is an easier and quicker way to find them? Thank you!

 

[mathbalarka]

I know how to find the roots with Horners method

You don't have to go that far. Try proving that this factors over integers completely by applying rational root theorem (integral root theorem for monic polynomials).

 

[wishmaster]

You don't have to go that far. Try proving that this factors over integers completely by applying rational root theorem (integral root theorem for monic polynomials).

I dont know how to do it......

 

[chisigma]

I have given polynomial: \(\displaystyle x^3-8x^2+19x-12\)

I know how to find the roots with Horners method,i am just wondering if there is an easier and quicker way to find them? Thank you!

The polynomial has degree 3, so that it exists at least one real root that can be found with the Newton-Raphson method...

 

[mathbalarka]

I dont know how to do it......

Rational root theorem - Wikipedia, the free encyclopedia

 

[wishmaster]

Rational root theorem - Wikipedia, the free encyclopedia

I will try it,but i cant see that this method can be shorter......again you have to test it it with Horners method. Or can you show me how to do it with my example?

 

[ZaidAlyafey]

Find the divisors of \(\displaystyle 12\) then divide that by the divisors of the leading term \(\displaystyle 1\) so we have \(\displaystyle \pm 1 , \pm 3 , \pm 4 , \pm 6 \pm 12\) as possible roots.

\(\displaystyle P(3) = 3^3-8(9)+19\times 3-12 = 27-72+57-12=0 \)

Hence \(\displaystyle x=3\) is a solution . The next step divide the cubic by \(\displaystyle (x-3)\) to get a polynomial of degree $2$.

 

[wishmaster]

Seems thats not the easier and faster method as Horners...or i cant see it!

 

[chisigma]

Seems thats not the easier and faster method as Horners...or i cant see it!

... not only neither easier nor faster... if the polynomial doesn't have rational roots [... the most probable case...] it fails!...

Kind regards

$\chi$ $\sigma$

 

[wishmaster]

... not only neither easier nor faster... if the polynomial doesn't have rational roots [... the most probable case...] it fails!...

Kind regards

$\chi$ $\sigma$

My question was about rational roots.....how to find them fast.

 

[mathbalarka]

neither easier nor faster

Horner's method has computational complexity $\mathcal{O}(n)$, whereas RRT exceeds that quite less often, as far as I can tell.

 

[wishmaster]

Horner's method has computational complexity $\mathcal{O}(n)$, whereas RRT exceeds that quite less often, as far as I can tell.

So i think the answer is: Deal with it as you can?

 

[mathbalarka]

Nope. I believe RRT is much faster at several cases than Horner, and that's what I answered above using computational complexity theory.