may i know what method should i use to find the roots of 3x^5 - 7x^4 + 11x^3 = 21x^2 - 22x + 8? can anybody get my confusion out from head please.

thank you

Printable View

- Jul 30th 2012, 10:18 AMrcsroots
may i know what method should i use to find the roots of 3x^5 - 7x^4 + 11x^3 = 21x^2 - 22x + 8? can anybody get my confusion out from head please.

thank you - Jul 30th 2012, 01:05 PMebainesRe: roots
Rearrange to get: 3x^5 - 7x^4+11x^3-21x^2+22x-8 = 0. By inspection it's clear that one root is x= 1. So divide through by (x-1) and you'll get a polynomial of 4th order, which it also turns out has a root of x=1. Continue until you get it to a 2nd order whose roots can be found "the usual way" using the quadratic equation.

- Jul 30th 2012, 07:37 PMbopbopRe: roots
Wolfram Alpha rearrangement and complex roots (if you need them)

You can solve quintics by factoring them into smaller parts (permutations or radicals). See Quintic Equation -- from Wolfram MathWorld. - Jul 30th 2012, 10:03 PMrcsRe: roots
thanks you bopbop ... can you show the solution bopbop?

thanks - Aug 1st 2012, 06:47 AMrcsRe: roots
i was using the synthetic division on this by x - 1... then there is remainder after i solved it in first value of divisor to be 1... meaning there is no roots but root ( 1 root ) . second question, is it considered a root when there is remainder of 16/ x-1?

- Aug 1st 2012, 07:05 AMebainesRe: roots
If you do the division of the polynomial by (x-1) three times you get:

$\displaystyle 3x^5 - 7x^4+11x^3-21x^2+22x-8 = (x-1)(3x^4-4x^3+7x^2-14x+8) = (x-1)^2(3x^3-x^2+6x-1) = (x-1)^3(3x^2+2x+1)$

So the roots are 1 and the two solutions to $\displaystyle 3x^2+2x+1=0$. - Aug 1st 2012, 08:24 AMBeanRe: roots
- Aug 1st 2012, 09:08 AMebainesRe: roots
- Aug 2nd 2012, 01:47 AMrcsRe: roots
thanks guys. God Bless