# Finding the roots of a high degree polynomial equation

1. Apr 16, 2013

### 1MileCrash

1. The problem statement, all variables and given/known data

y(6) - 3y(4) + 3y''-y = 0

2. Relevant equations

3. The attempt at a solution

The characteristic equation of that differential equation is:

r^6 - 3r^4 + 3r^2 - r = 0

But how am I expected to solve such a high degree polynomial (and thus the DE?)

2. Apr 16, 2013

### Bacle2

First, notice you can rewrite you char. equation as:

r(r^5-3r^3+3r-1)=0

Then notice too, that, inside your parenthesis: 1-3+3-1=0. What does this tell you?

3. Apr 16, 2013

### tiny-tim

Hi 1MileCrash!

(try using the X2 button just above the Reply box )
nooo

r6 - 3r4 + 3r2 - 1 = 0

4. Apr 16, 2013

### SammyS

Staff Emeritus
Following tiny-tim's corrected characteristic equation:
r6 - 3r4 + 3r2 - 1 = 0​

Expand (a - b)3 .

5. Apr 16, 2013

### epenguin

I don't know at what point certain things are supposed to be familiar but I would say pretty early there is a rather familiar pattern to be discerned in that last formula.

6. Apr 16, 2013

### HallsofIvy

$r^6- 3r^4+ 3r^2- 1$ has only even powers of r. Let $x= r^2$ and that becomes $x^3- 3x^2+ 3x- 1$. And, as SammyS suggests, that is $(x- 1)^3$.

7. Apr 16, 2013

### Bacle2

Worse comes to worse and you cannot see this pattern, it always makes sense to try for the simple roots,like 0,1 and -1 . Checking for 1 as a root comes down to adding the coefficients and seeing if the sum is zero; similar for -1.

8. Apr 17, 2013

### HallsofIvy

Let me point out that Bacle2 is not just choosing "simple roots" at random. Since the leading coefficient is 1 and the constant term is 1, by the "rational root theorem" the only possible rational number roots are 1 and -1. (I don't know why he mentions "0".)

9. Apr 17, 2013

### Bacle2

Right, my bad. I thought the characteristic equation had no constant term.

10. Apr 17, 2013

### LCKurtz

All this help and the OP is nowhere in sight.