# Simple polynomial question

1. Dec 8, 2006

### ACLerok

Hi, I was just reading up on some material for one of my exams and came across something. I realize this may seem like a trivial, obvious question but:
What is the easiest way to find the solutions of the polynomial (for example):
(x^2) - (1/6)x - 1/6 = 0
Other than guessing and testing for the correct values of x, what is the quickest and most efficient method of solving for x when the coefficients of each term is a fraction rather than a whole number? Thanks in advance.

2. Dec 8, 2006

### Gib Z

well you just let your fraction, 1/6, equate to b, as the equation was a quadratic of the form $ax^2 + bx + c = 0$. Then substitute your values for a, b and c into the quadratic formula, $$x=\frac {-b(+-)(b^2 - 4ac)^0.5}{2a}$$ If you mean in general, you it is quite hard to solve equations of degree 3 and 4 analytically, and after 5 it can only be approximated using methods such as Newtons Method.

3. Dec 8, 2006

### Edgardo

Given the equation

$$x^2 + px + q = 0$$

the solutions are:

$$x_{1} = - \frac{p}{2} - \sqrt{ \left( \frac{p}{2} \right)^2-q }$$

and

$$x_{2} = - \frac{p}{2} + \sqrt{ \left( \frac{p}{2} \right)^2-q }$$

So in your example:

p=-1/6 and q=-1/6

Last edited: Dec 8, 2006
4. Dec 8, 2006

Thanks alot!

5. Dec 8, 2006

### ACLerok

one more thing: if im trying to find the solutions if the polynomial is of order 3, for example:
x^3 + 0.5x^2 - 0.25x - 0.125 = 0

Do I add 0.125 to both sides then factor an x or do must I do something else? Thanks

6. Dec 8, 2006

7. Dec 9, 2006

### ACLerok

Wow what a long and involved process
There is no other process of finding them? oh well
Thanks

8. Dec 10, 2006

### nworm

You can try to find (adjust) one solution ($$x=0.5$$). After that you can divide $$x^3 + 0.5x^2 - 0.25x - 0.125$$ by $$x-0.5$$.
The result is $$x^2+x+0.25$$.
So $$x^3 + 0.5x^2 - 0.25x - 0.125=(x-0.5)(x^2+x+0.25)=0$$
You already can solve it.

Last edited: Dec 10, 2006
9. Dec 10, 2006

### arildno

Not all problems easily stated are easily solved.

Most happen to be insoluble, I'm afraid.

10. Dec 11, 2006

### nworm

arildno
I don't think that ACLerock solve very difficult problems.
It is possible to apply more easy methods when solving home and exam tasks.
(for example the method from my post).

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