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Roots of the polynomial

wishmaster

Active member
Oct 11, 2013
211
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

Well-known member
MHB Math Helper
Mar 22, 2013
573
wishmaster said:
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

Active member
Oct 11, 2013
211
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

Well-known member
Feb 13, 2012
1,704
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...
 

wishmaster

Active member
Oct 11, 2013
211

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Active member
Oct 11, 2013
211
Seems thats not the easier and faster method as Horners...or i cant see it! :D
 

chisigma

Well-known member
Feb 13, 2012
1,704
Seems thats not the easier and faster method as Horners...or i cant see it! :D
... not only neither easier nor faster... if the polynomial doesn't have rational roots [... the most probable case...] it fails!(Tmi)...

Kind regards

$\chi$ $\sigma$
 

wishmaster

Active member
Oct 11, 2013
211
... not only neither easier nor faster... if the polynomial doesn't have rational roots [... the most probable case...] it fails!(Tmi)...

Kind regards

$\chi$ $\sigma$
My question was about rational roots.....how to find them fast. (Mmm)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
chisigma said:
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

Active member
Oct 11, 2013
211
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

Well-known member
MHB Math Helper
Mar 22, 2013
573
Nope. I believe RRT is much faster at several cases than Horner, and that's what I answered above using computational complexity theory.