# Efficient Solutions for Fractional Polynomial Equations

• ACLerok
In summary, the easiest way to find solutions for a polynomial with fractions as coefficients is to use the quadratic formula, x=\frac {-b(+-)(b^2 - 4ac)^0.5}{2a}. For polynomials of higher degrees, there are methods such as Cardano's method for cubic equations, but they can be long and involved. It is also possible to try to find one solution and then divide the polynomial by that solution to simplify the problem. However, not all problems are easily solvable.
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.

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.

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

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

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

Last edited:
Thanks alot!

one more thing: if I am 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

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

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 $$x^3 + 0.5x^2 - 0.25x - 0.125=(x-0.5)(x^2+x+0.25)=0$$

Last edited:
ACLerok said:
Wow what a long and involved process
There is no other process of finding them? oh well
Thanks

Not all problems easily stated are easily solved.

Most happen to be insoluble, I'm afraid.

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).

## 1. What is a simple polynomial?

A simple polynomial is an algebraic expression made up of variables, coefficients, and mathematical operations such as addition, subtraction, multiplication, and exponentiation. It does not involve any complex mathematical functions or operations.

## 2. How do you solve a simple polynomial equation?

To solve a simple polynomial equation, you need to simplify the expression by combining like terms and using the basic rules of algebra. Then, isolate the variable on one side of the equation by performing inverse operations, and you will get the solution.

## 3. What is the degree of a simple polynomial?

The degree of a simple polynomial is the highest exponent of the variable in the expression. It represents the number of terms in the polynomial and determines the complexity of the equation.

## 4. Can a simple polynomial have more than one variable?

Yes, a simple polynomial can have more than one variable. In fact, a polynomial can have any number of variables, but they must be raised to whole number exponents.

## 5. What are some real-life applications of simple polynomials?

Simple polynomials are used in various fields of science and engineering, such as physics, chemistry, and economics, to model real-world phenomena and solve problems. They are also used in computer programming and data analysis for curve fitting and data representation.

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