# Homework Help: Solving Polynomials

1. Jan 20, 2007

### DieCommie

1. The problem statement, all variables and given/known data
Solve the following equations in terms of complex numbers without a calculator.

d) x^3-x^2-x-2=0

e) x^4-2x^3+3x^2-2x+2=0

2. Relevant equations
?

3. The attempt at a solution
I dont know where to begin.

I dont think I was taught this in college algebra unfortunatly. I know there is some god awful formula that will allow me to solve d), but something tells me thats not what the teacher has in mind. I have no clue where to even begin solving e).

I know Im supposed to make an attempt, all I have done on paper so far is factor out x and rearrange things.

Any hints would be appreciated, thx

2. Jan 20, 2007

### Hurkyl

Staff Emeritus
(d) should be easy. Do you know any interesting theorems about roots of polynomials? There are 4 things you should try immediately, and one of them works.

(e) is an exercise in factoring. That polynomial can be factored. Can you show that it has no (rational) linear factors? If it has no linear factors, then what sort of factors must it have? And what must they look like?

3. Jan 20, 2007

### DieCommie

d)I know (or I think) that the number of solutions equals the degree of the polynomial. So I expect three solutions. Four things I could try? I could try the formula... I could try to factor it by trial and error. I could guess some values of x and plug them in which may lead me to the correct answer, or tell me the ballpark.

e)"Can I show that it has no linear factors?" No, I cannot. How would I show that? Maybe try to factor out a (x+a) and see how that works?

"If it has no linear factors, then what sort of factors must it have? And what must they look like?" Well, assuming it has no linear factors, then the factors would have to be of the form (x+a)^n where n>1?

4. Jan 20, 2007

### Hurkyl

Staff Emeritus
There is a theorem which, when applied to this particular polynomial, gives you a list of exactly four things that are worth guessing....

If (x+a)^n was a factor, then (x+a) is a factor, so your polynomial would have a linear factor.

5. Jan 20, 2007

### HallsofIvy

Hurkyl is talking about the "rational root theorem": If x= m/n, with m and n integers with no common factor, is a rational number solution to $a_nx^n+ ...+ a_0= 0$ where all coefficients are integers, then m must divide $a_0$ and n must divide $a_n$. In your case, $a_n$= 1 and $a_0= 2$. There are exactly 2 integers that divide 1 and 4 integers that divide 2 and it turns out there are exactly 4 possible values for m/n. Try them.

No one is saying that all those are roots. In fact, it is quite possible that a polynomial equation has no rational roots but it is worth trying. While there exist formula for solving general cubic and quartic equations, they are so complicated, it is much better searching for rational roots. Of course, if you can find a root m/n, dividing the polynomial by x- m/n reduces the degree. Perhaps you can find enough to reduce the problems to quadratic equations which you can then solve with the quadratic formula.

As for your question about "linear factors", EVERY polynomial can be factored into linear or quadratic factors that cannot be further factored. The quadratic x2- 1 can be factored into (x- 1)(x+ 1) so that x2- 1= 0 has roots 1 and -1. An example of a quadratic factor that cannot be further factored is x2+ 1. What does that tell you about the roots of x2+ 1= 0?

Last edited by a moderator: Jan 20, 2007
6. Jan 21, 2007

### DieCommie

I have never heard of "rational root theorem". Im also a little confused as to what the term "m" is.

I am going to go ahead and use the formulas to solve these:yuck:

7. Jan 21, 2007

### matt grime

Don't. Try to work through it with what was written above.

You can prove the rational root theorem for yourself: it is straight forward. You suppose there is a rational root, and call it n/m with n and m integers with no common factors. This gives you easy relations between m and n.

If you don't like that then the first thing you should try to do when given some polynomial is just plug in 0,1,2,-1, and -2. They're not trying to make you work very hard.

8. Jan 21, 2007

### Gib Z

Your actually going to use the forumulaes? You have gone insane. The equations are massively long, and theres 3 seperate ones to solve for the cubic. Even longer, and 4 to solve for the quartic. You must be really scared of this theorem...