1. ## finding the roots

what is the easiest way of finding the roots of these kind of equations (except guessing):

x^4+x^3-7x^2-x+6=0

or

x^5-2x^4+2x^3-2x^2+x=0

you can tell rigorously or sth easy. it's OK.

2. a) $x^4+x^3-7x^2-x+6=0$

Find integer roots. $x_1=1, \ x_2=-1$

b) $x(x^4-2x^3+2x^2-2x+1)=0$

$x_1=0$ and $x^4-2x^3+2x^2-2x+1=0$ is a reciprocal equation.

3. Start by identifying that:

x^4 + x^3 -7x^2 -x +6 = 0
So x(x^3 + x^2 - 7x -1) = -6
--> x^2(x^2 + x -7) = -5
--> x^3(x +1) = 2
--> x^4(1) = 1

Therefore, your first x is going to be 1, and your second x is going to be -1.

Next, use these roots to synthetically divide the equation down into something simpler or merely factor out the roots (x+1) and (x-1).

You are left with (x+1)(x-1)(x^2 + x -6) which simplifies to (x+1)(x-1)(x+3)(x-2).

4. Originally Posted by eyke
what is the easiest way of finding the roots of these kind of equations (except guessing):

x^4+x^3-7x^2-x+6=0

x^5-2x^4+2x^3-2x^2+x=0
After looking at the graph in your graphing calculator, you can start by applying the Rational Roots Test to get a list of possible rational (fractional or whole-number) zeroes.

Then you can try them (picking the more-likely options from the list by comparing with the calculator's picture) in synthetic division. Whichever values give you a "zero" remainder are valid roots.

Then you apply the same process to whatever is remaining. The steps are illustrated here. :wink: