# roots

• Jul 30th 2012, 10:18 AM
rcs
roots
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 PM
ebaines
Re: 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 PM
bopbop
Re: 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 PM
rcs
Re: roots
thanks you bopbop ... can you show the solution bopbop?

thanks
• Aug 1st 2012, 06:47 AM
rcs
Re: 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 AM
ebaines
Re: roots
If you do the division of the polynomial by (x-1) three times you get:

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

So the roots are 1 and the two solutions to $3x^2+2x+1=0$.
• Aug 1st 2012, 08:24 AM
Bean
Re: roots
Quote:

Originally Posted by ebaines
If you do the division of the polynomial by (x-1) three times you get:

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

So the roots are 1 and the two solutions to $3x^2+2x+1=0$.

You lost an 8 in there. The answer should be $(x-1)^3(3x^2+2x+8)$.

$(3x^2+2x+8)$ is irreducible.
• Aug 1st 2012, 09:08 AM
ebaines
Re: roots
Quote:

Originally Posted by Bean
You lost an 8 in there. The answer should be $(x-1)^3(3x^2+2x+8)$.

Right - typo on my part, good catch.

Quote:

Originally Posted by Bean
$(3x^2+2x+8)$ is irreducible.

Apply the quadratic formula, and you find $3x^2 + 2x + 8 = 3(x + \frac 1 3 + \frac 2 3 \sqrt{23} i )(x + \frac 1 3 - \frac 2 3 \sqrt{23} i )$
• Aug 2nd 2012, 01:47 AM
rcs
Re: roots
thanks guys. God Bless