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
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.
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.
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?
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 $\displaystyle 3x^2+2x+1=0$.