How do you solve a 4th degree equation?

[nmbr42]

Homework Statement

From the higher order linear homogeneous differential equation, I get the characteristic equation:
r^4 + r^3 -7r^2 -r + 6 = 0
solve for r.

Homework Equations

how do you do this?

The Attempt at a Solution

Even if I factor out the r from the first 4 terms on the LH, i get no where. I wish in high school they went over this stuff, but they seriously didn't

 

[SammyS]

Homework Statement

From the higher order linear homogeneous differential equation, I get the characteristic equation:
r^4 + r^3 -7r^2 -r + 6 = 0
solve for r.

Homework Equations

how do you do this?

The Attempt at a Solution

Even if I factor out the r from the first 4 terms on the LH, i get no where. I wish in high school they went over this stuff, but they seriously didn't

It's fairly easy to see that r = 1 is a solution.

Almost as easy to see that r = -1 is also a solution.

 

[Simon Bridge]

nmbr42: the usual approach is to guess, using your experience of how the equations work, then use long division to remove the guessed factors. It is usually taught in junior High School. Possibly you were sick that day or more interested in other things.

How easy it is to realize that 1 and -1 are roots does kinda depend on your experience with these things... but you do learn how to make good guesses as you become familiar with the way polynomials behave.

In a pinch you can use Newton/Raphson to get the first root, or just use a computer to plot the graph :) and there are general solutions online.

 

[HallsofIvy]

One important guide to making "guesses" is the "rational root theorem".

"Any rational number, satisfying $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$, is of the form m/n where m evenly divides the "constant term", $a_0$ and n evenly divides the "leading coefficient", $a_n$."

Of course, it is always possible that a polynomial equation does not have any rational roots but we can tell that if this equation has rational roots, they must be integers (because the leading coefficent is 1 and only 1 divides 1) and must be 1, -1, 2, -2, 3, -3, 6, or -6 (because the constant term is 6 and only those numbers divide 6).

 

[nmbr42]

I actually used this link to help me since Simon Bridge mentioned about long division:
http://www.purplemath.com/modules/polydiv3.htm

quite simple straight forward method.
the roots are -3,2,1, and -1.
Since -1 and 1 are guesses that actually worked, I first divided the equation by (r-1) then (r+1). Then at last you get the equation
r^2 + r -6 =0

 

[Simon Bridge]

no worries... it's all good and glad we could help :)