# Roots of the polynomial

#### wishmaster

##### Active member
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
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
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
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...

MHB Math Helper

#### wishmaster

##### Active member
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

##### Well-known member
MHB Math Helper
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
Seems thats not the easier and faster method as Horners...or i cant see it! #### chisigma

##### Well-known member
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

##### Active member
... 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

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